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Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 11 Straight Lines Ex 11(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(a)

Question 1.
Find the distance between the following pairs of points.
(i) (3, 4), (-2, 1);
Solution:
Distance between points (3, 4) and (-2, 1) is
\(\sqrt{(3+2)^2+(4-1)^2}=\sqrt{25+9}=\sqrt{34}\)

(ii) (-1, 0), (5, 3)
Solution:
The distance between the points (-1, 0) and (5, 3) is
\(\sqrt{(-1-5)^2+(0-3)^2}\)
= \(\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\)

Question 2.
If the distance between points (3, a) and (6, 1) is 5, find the value of a.
Solution:
Distance between the points. (3, a) and (6, 1) is
\(\sqrt{(3-6)^2+(a-1)^2}=\sqrt{9+(a-1)^2}\)
∴ \(\sqrt{9+(a-1)^2}=5\)
or, 9 + (a – 1)² = 25
or, (a – 1)² = 16
or, a – 1 = ± 4
a = 1 ± 4 = 5 or, – 3

Question 3.
Find the coordinate of the points which divides the line segment joining the points A (4, 6), B (-3, 1) in the ratio 2: 3 internally. Find also the coordinates of the point which divides \(\overline{\mathbf{A B}}\) in the same ratio externally.
Solution:

Question 4.
Find the coordinates of the mid-point of the following pairs of points.
(i) (-7, 3), (8, -4);
Solution:
Mid-point of the line segment joining the points (-7, 3) and (8, -4) are \(\left(\frac{-7+8}{2}, \frac{3-4}{2}\right)=\left(\frac{1}{2},-\frac{1}{2}\right)\)

(ii) (\(\frac{3}{4}\), -2), (\(\frac{-5}{2}\), 1)
Solution:
Mid-point of the line segment joining the points. (\(\frac{3}{4}\), -2) and (\(\frac{-5}{2}\), 1) is,
\(\left(\frac{\frac{3}{4}-\frac{5}{2}}{2}, \frac{-2+1}{2}\right)=\left(\frac{-7}{8}, \frac{-1}{2}\right)\)

Question 5.
Find the area of the triangle whose vertices are (1, 2), (3, 4) (\(\frac{1}{2}\), \(\frac{1}{4}\))
Solution:
Area of the triangle whose vertices are (1, 2), (3, 4) and (\(\frac{1}{2}\), \(\frac{1}{4}\)) is

Question 6.
If the area of the triangle with vertices (0, 0), (1, 0), (0, a) is 10 units, find the value of a.
Solution:
Area of the triangle with vertices (0, 0),(1,0), (0, a), is \(\frac{1}{2}\) × 1 × a = \(\frac{a}{2}\)
∴ \(\frac{a}{2}\) = 10 or a = 20

Question 7.
Find the value of a so that the points (1, 4), (2, 7), (3, a) are collinear.
Solution:
As points (1, 4), (2, 7), (3, a) are collinear, we have the area of the triangle with vertices (1, 4), (2, 7), and (3, a) is zero.
∴ \(\frac{1}{2}\) {1(7 – a) + 2(a – 4) + 3 (4 – 7)} = 0
or, 7 – a + 2a – 8 + 12 – 21 =0
⇒ a = 10

Question 8.
Find the slope of the lines whose inclinations are given.
(i) 30°
Solution:
The slope of the line whose inclination is 30°.
tan 30° = \(\frac{1}{\sqrt{3}}\)

(ii) 45°
Solution:
Slope = tan 45° = +1

(iii) 60°
Solution:
Slope = tan 60° = √3

(iv) 135°
Solution:
Slope = tan 135° = – 1

Question 9.
Find the inclination of the lines whose slopes are given below.
(i) \(\frac{1}{\sqrt{3}}\)
Solution:
The slope of the line is \(\frac{1}{\sqrt{3}}\)
∴ tan θ = \(\frac{1}{\sqrt{3}}\) or, θ = 30°
∴ The inclination of the line is 30°

(ii) 1
Solution:
Slope = 1 = tan 45°
∴ The inclination of the line is 45°.

(iii) √3
Solution:
Slope = √3 = tan 60°  ∴ θ = 60°
∴ Inclination = 60°

(iv) – 1
Solution:
Slope = – 1 = tan 135°
∴ Inclination = 135°

Question 10.
Find the angle between the pair of lines whose slopes are ;
(i) \(\frac{1}{\sqrt{3}}\), 1
Solution:

(ii) √3, -1
Solution:

Question 11.
(a) Show that the points (0, -1), (-2, 3), (6, 7), and (8, 3) are vertices of a rectangle.
Solution:


∴ The opposite sides are equal and two consecutive sides are perpendicular. So it is a rectangle.

(b) Show that the points (1, 1), (-1, -1), and (-√3, √3) are the vertices of an equilateral triangle.
Solution:

Question 12.
Find the coordinates of the point P(x, y) which is equidistant from (0, 0), (32, 10), and (42, 0).
Solution:

Question 13.
If the points (x, y) are equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

Question 14.
The coordinate of the vertices of a triangle are (α₁, β₁), (α₂, β₂), and (α₃, β₃). Prove that the coordinates of its centroid is \(\left(\frac{\alpha_1+\alpha_2+\alpha_3}{3}, \frac{\beta_1+\beta_2+\beta_3}{3}\right)\)
Solution:

Question 15.
Two vertices of a triangle are (0, -4) and  (6, 0). If the medians meet at the point (2, 0), find the coordinates of the third vertex.
Solution:

∴ \(\frac{6+x}{3}\) = 2, \(\frac{-4+y}{3}\) = 0
⇒ x = 0, y = 4
∴ The coordinates of the 3rd vertex are (0, 4).

Question 16.
If the point (0, 4) divides the line segment joining(-4, 10) and (2, 1) internally, find the point which divides it externally in these same ratios.
Solution:

Question 17.
Find the ratios in which the line segment joining (-2, -3) arid (5, 4) is divided by the coordinate axes and hence find the coordinates of these points.
Solution:

Question 18.
In a triangle, one of the vertices is at (2, 5) and the centroid of the triangle is at (-1, 1). Find the coordinates of the midpoint of the side opposite to the given angular point.
Solution:

Question 19.
Find the coordinates of the vertices of a triangle whose sides have midpoints at (2, 1), (-1, 3), and (-2, 5).
Solution:

∴ x₂ + x₃ – 4 or, x₂ – 4 – 3 = 1
∴ x₁ = – 4 – x₂ = -4 – 1 = -5
Similarly y₁ + y₂ + y₃ = 5 + 1 + 3 = 9
As y₁ + y₂ = 10
we have y₃ = 9 – 10 = – 1
Again y₁ + y₃ = 6
or, y₁ = 6 – y₃ = 6 + 1 = 7
and y₂ = 10 – y₁ = 10 – 7 = 3
∴ The coordinates of A, B, and C are (-5, 7), (1, 3), and (3, -1).

Question 20.
If the vertices of a triangle have their coordinates given by rational numbers, prove that the triangle cannot be equilateral.
Solution:
Let us choose the contradiction method. Let the triangle is equilateral if the co¬ ordinate of the vertices is rational numbers.
Let ABC be an equilateral triangle with vertices A (a, 0), B (a, 0), and C (b, c) where a, b, c are rational.

⇒ a² = b² + c² = \(\frac{a^2}{4}\) + c²
⇒ c² =  a² – \(\frac{a^2}{4}\) = \(\frac{3a^2}{4}\) ⇒ c = \(\frac{\sqrt{3}}{2}\) a     ….(2)
Now b = \(\frac{a}{2}\), c = \(\frac{\sqrt{3}}{2}\) a
If a is rational then b is rational but c is irrational, i.e., the coordinates of the vertices are not rational, which contradicts the assumption.
Hence assumption is wrong.
So the triangle cannot be equilateral if the coordinate of the vertices is rational numbers.

Question 21.
Prove that the area of any triangle is equal to four times the area of the triangle formed by joining the midpoints of its sides.
Solution:

∴ The area of triangle ABC is four times the area of triangle DEF. (Proved)

Question 22.
Find the condition that the point (x, y) may lie on the line joining (1, 2) and (5, -3).
Solution:

∴ As points A, B, and C are collinear, we have the area of the triangle ABC as 0.
∴ \(\frac{1}{2}\) {1(-3 – y) + 5(y – 2) + x(2 + 3)} = 0
or, – 3 – y + 5y – 10 + 5x = 0
or, 5x + 4y = 13

Question 23.
Show that the three distinct points (a², a), (b², b), and (c², c) can never be collinear.
Solution:
Area of the triangle with vertices (a², a), (b², b) , and (c², c) is
\(\frac{1}{2}\) {a²(b – c) + b²(c – a) c²(a – b)}
= (a – b)(b – c)(a – c)
which is never equal to zero except when a = b = c, hence the points are not collinear.

Question 24.
If A, B, and C are points (-1, 2), (3, 1), and (-2, -3) respectively, then show that the points which divide BC, CA, and AB in the ratios (1: 3), (4: 3) and (-9: 4) respectively are collinear.
Solution:
Let the points P, Q, and R divides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\), in \(\overline{\mathrm{AB}}\) the ratio 1: 3, 4: 3 and -9: 4

Question 25.
Prove analytically :
(a) The line segment joining the midpoints of two sides of a triangle is parallel to the third and half of its length.

Solution:
Let the coordinates of the triangle ABC be (x₁, y₁), (x₂, y₂) and (x₃, y₃)
The points D and E are the midpoints of the sides \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)

(b) The altitudes of a triangle are concurrent.
Solution:

(c) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:

(d) An angle in a semicircle is a right angle.
Solution:

∴ The angle in a semicircle is a right angle. (Proved)

Odisha State Board BSE Odisha 10th Class Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(c) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Algebra Chapter 1 ସରଳ ସହସମୀକରଣ Ex 1(c)

Question 1.
ଦୁଇଟି ସଂଖ୍ୟାର ଯୋଗଫଳ 137 ଓ ସେମାନଙ୍କର ବିୟୋଗଫଳ 43। ତେବେ ସଂଖ୍ୟାଦ୍ୱୟ ନିରୂପଣ କର ।
ସମାଧାନ:
ମନେକର ସଂଖ୍ୟାଦ୍ଵୟ x ଓ y ।
ପ୍ରଶ୍ନନୁସାରେ x + y = 137 ……..(i) ଏବଂ x – y = 43 …….(ii)
ସମୀକରଣ (i) ଓ (ii) କୁ ପ୍ରୟୋଗକଲେ,

x ର ମାନ ସମୀକରଣ (i) ରେ ସ୍ଥାପନ କଲେ, x + y = 137
⇒ 90 + y = 137 ⇒ y = 137 – 90 ⇒ y = 47
∴ ସଂଖ୍ୟାଦ୍ଵୟ 90 ଓ 47।

Question 2.
ସମବାହୁ ତ୍ରିଭୁଜର ବାହୁ ତ୍ରୟର ଦୈର୍ଘ୍ୟ x + 4 ସେ.ମି., 4x – y ସେ.ମି. ଓ y + 2 ସେ.ମି. ହେଲେ ବାହୁର ଦୈର୍ଘ୍ୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ସମବାହୁ ତ୍ରିଭୁଜର ବାହୂତ୍ରୟର ଦୈର୍ଘ୍ୟ ସମାନ । ଅର୍ଥାତ୍ x + 4 = 4x – y = y + 2
⇒ x + 4 = y + 2 = x – y = 2 – 4
⇒ x – y = -2 … (i)
ପୁନଶ୍ଚ 4x – y = y + 2 ⇒ 4x – y – y = 2 ⇒ 4x – 2y = 2
⇒ 2(2x – y) = 2 ⇒ 2x – y = 1 …… (ii)
ସମୀକରଣ (ii)ରୁ ସମୀକରଣ (i)କୁ ବିୟୋଗ କଲେ,

x ର ମାନ ସମୀକରଣ (i) ରେ ସ୍ଥାପନ କଲେ, 3 – y = – 2
⇒ – y = -2 – 3 = -5 ⇒ y = 5
∴ ସମବାହୁ ତ୍ରିଭୁଜର ଏକ ବାହୁର ଦୈର୍ଘ୍ୟ = y + 2 = 5 + 2 = 7 ସେ.ମି.

Question 3.
ABCD ଅ।ୟତକ୍ଷେତ୍ରର AB = 3x + y ସେ.ମି, BC = 3x + 2 ସେ.ମି, CD = 3y – 2x ସେ.ମି, ଓ DA = y + 3 ସେ.ମି. ହେଲେ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ନିରୂପଣ କର ।
ସମାଧାନ :
ଆୟତଚିତ୍ରର ବିପରୀତ ବାହୁମାନଙ୍କର ଦୈର୍ଘ୍ୟ ପରସ୍ପର ସମାନ ।
ତେଣୁ ABCD ଆୟତଚିତ୍ରରେ AB = CD ଏବଂ BC = AD ହେବ ।
AB = CD ହେଲେ, 3x + y = 3y – 2x ⇒ 3x + 2x = 3y – y ⇒ 5x – 2y=0 (i)
ସେହିପରି BC = AD ହେଲେ, 3x + 2 = y + 3
⇒ 3x – y = 3 – 2 ⇒ 3x – y =1 …..(ii)

‘x’ ର ମାନ ସମୀକରଣ (ii) ରେ ପ୍ରୟୋଗ କଲେ, 3 × 2 – y = 1 ⇒ 6 – 1 = y ⇒ y = 5
∴ AB = 3x + y = 3 × 2 + 5 = 11 ସେ.ମି
BC = 3x + 2 = 3 × 2 + 2 = 8 ସେ.ମି
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = AB × BC = 11 × 8 = 88

Question 4.
ଦୁଇ ଅଙ୍କ ବିଶିଷ୍ଟ ଗୋଟିଏ ସଂଖ୍ୟା, ତାହାର ଅଙ୍କଦ୍ଵୟର ଯୋଗଫଳର 4 ଗୁଣ । କିନ୍ତୁ ସଂଖ୍ୟାଟିରେ 18 ଯୋଗ କଲେ ଅଙ୍କଦ୍ଵୟର ସ୍ଥାନ ବଦଳିଯାଏ । ତେବେ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାଟିର ଦଶକ ଏବଂ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କଦ୍ଵୟ ଯଥାକ୍ରମେ x ଏବଂ y l .:. ସଂଖ୍ୟାଟି = 10x + y
ପ୍ରଶ୍ନନୁସାରେ, ସଂଖ୍ୟାଟି ତାହାର ଅଙ୍କ ଦ୍ବୟର ଯୋଗଫଳର 4 ଗୁଣ ।
⇒ 10x + y = 4(x + y) ⇒ 10x + y = 4x + 4y ⇒ 10x – 4x + y – 4y = 0
⇒ 6x – 3y = 0 ⇒ 3(2x – y) = 0 … (i)
⇒2x – y = 0
ପ୍ରଶ୍ନନୁସାରେ ସଂଖ୍ୟାଟିରେ 18 ଯୋଗକଲେ ଅଙ୍କ ଦୁଇଟିର ସ୍ଥାନ ବଦଳିଯାଏ ।
⇒ 10x + y + 18 = 10y + x ⇒ 10x – x + y – 10y = -18
⇒ 9(x- y) = -18 ⇒ x – y = \(\frac{-18}{9}=-2\)

x ର ମାନ ସମୀକରଣ (i)ରେ ସଂସ୍ଥାପନ କଲେ, 2x – y = 0
⇒ 4 – y = 0 ⇒ y=4
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 2 + 4 = 24

Question 5.
ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ଗୋଟିଏ ସଂଖ୍ୟା ଓ ତାହାର ଅଙ୍କଦ୍ଵୟର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ଯେଉଁ ସଂଖ୍ୟା ମିଳିବ, ସେ ଦୁହିଁଙ୍କର ଯୋଗଫଳ 99 ଓ ଅଙ୍କଦ୍ୱୟର ଅନ୍ତର 3 ହେଲେ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଦଶକ ଓ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ ଦ୍ବୟ ଯଥାକ୍ରମେ x ଓ y
∴ ସଂଖ୍ୟାଟି = 10x + y
ଅଙ୍କଦ୍ୱୟର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ସଂଖ୍ୟାଟି ହେବ = 10y + x
ପ୍ରଶ୍ନନୁସାରେ, 10x + y + 10y + x = 99
⇒ 11x + 11y = 99 ⇒ x + y = \(\frac{99}{11}=9\) ………(i)
ଅଙ୍କଦ୍ୱୟର ଅନ୍ତର 3 ।
ଅର୍ଥାତ୍ x – y = 3 ବା y – x = 3 … (ii)
ଯଦି x – y = 3 ହୁଏ,
∴ x = \(\frac{x+y+x-y}{2}=\frac{9+3}{2}=\frac{12}{2}=6\)
y = 9 – x = 9 – 6 = 3
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 6 + 3 = 63
ଯଦି ଆମେ ଦ୍ବିତୀୟ ସମୀକରଣକୁ y – x = 3 ନେବା ତେବେ ସଂଖ୍ୟାଟି 36 ହେବ ।
∴ ସଂଖ୍ୟାଟି 63 ବା 36 ।

Question 6.
ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି, ସେମାନଙ୍କ ବିୟୋଗଫଳର 4 ଗୁଣ ଏବଂ ସଂଖ୍ୟା ଦୁଇଟିର ଯୋଗଫଳ 8 । ତେବେ ସଂଖ୍ୟା ଦୁଇଟି କେତେ ?
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟାଦ୍ଵୟ x ଓ y
ପ୍ରଶ୍ନନୁସାରେ, x + y = 4 (x – y)
⇒ 4x – 4y – x – y = 0 ⇒ 3x – 5y = 0 … (i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ x + y = 8 …… (ii)

x ର ମାନ ସମୀକରଣ (ii)ରେ ସଂସ୍ଥାପନ କଲେ, x + y = 8
⇒ y = 8 – x = 8 – 5 = 3
∴ ସଂଖ୍ୟାଦ୍ୱୟ 5 ଓ 3 ।

Question 7.
ଦୁଇଅଙ୍କ ବିଶିଷ୍ଟ ସଂଖ୍ୟାର ଅଙ୍କମାନଙ୍କର ସମଷ୍ଟି 10; କିନ୍ତୁ ଅଙ୍କଗୁଡ଼ିକର ସ୍ଥାନ ବଦଳାଇ ଲେଖୁଲେ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି ମୂଳ ସଂଖ୍ୟାର ଦୁଇ ଗୁଣରୁ 1 ଊଣା ହୁଏ, ସଂଖ୍ୟାଟି ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟାଟିର ଦଶକ ସ୍ଥାନୀୟ ଅଙ୍କ ଏବଂ ଏକକ ସ୍ଥାନୀୟ ଅଙ୍କ ଯଥାକ୍ରମେ x ଏବଂ y ।
∴ ସଂଖ୍ୟାଟି 10x + y ହେବ ।
ଅଙ୍କଗୁଡ଼ିକର ସ୍ଥାନ ବଦଳିଗଲେ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି 10y + x ହେବ ।
ପ୍ରଶ୍ନନୁସାରେ, ଅଙ୍କଗୁଡ଼ିକର ଯୋଗଫଳ 10 । x + y = 10 … (i)
ପୁନଶ୍ଚ ଉତ୍ପନ୍ନ ସଂଖ୍ୟାଟି, ମୂଳ ସଂଖ୍ୟାର ଦୁଇଗୁଣରୁ 1 କମ୍ ।
ଅର୍ଥାତ୍ 10y + x = 2(10x + y) – 1 = 10y + x = 20x + 2y – 1
⇒ 8y – 19x = -1 ⇒ 19x – 8y = 1 … (ii)

‘x’ ର, ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, 3 + y = 10 ⇒ y = 10 -3 = 7
∴ ସଂଖ୍ୟାଟି = 10x + y = 10 × 3 + 7 = 37

Question 8.
ଦୁଇଟି ସଂଖ୍ୟା ମଧ୍ୟରୁ ପ୍ରଥମଟିର 3 ଗୁଣରୁ ଦ୍ବିତୀୟଟିର 2 ଗୁଣ ବିୟୋଗ କଲେ ବିୟୋଗଫଳ 2 ହେବ ଏବଂ ଦ୍ଵିତୀୟଟିରେ 7 ଯୋଗ କଲେ ଯୋଗଫଳ ପ୍ରଥମଟିର 2 ଗୁଣ ହେବ। ସଂଖ୍ୟାଦ୍ଵୟ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ପ୍ରଥମ ସଂଖ୍ୟା ଓ ଦ୍ୱିତୀୟ ସଂଖ୍ୟା ଯଥାକ୍ରମେ x ଏବଂ y ହେଉ ।
ପ୍ରଥମଟିର 3 ଗୁଣରୁ ଦ୍ବିତୀୟଟିର 2 ଗୁଣ ବିୟୋଗକଲେ ବିୟୋଗଫଳ 2 ହେବ ।
ଅର୍ଥାତ୍ 3x – 2y = 2 …(i)
ସେହିପରି ଦ୍ବିତୀୟ ସଂଖ୍ୟାରେ 7 ଯୋଗକଲେ, ଯୋଗଫଳ ପ୍ରଥମ ସଂଖ୍ୟାର 2 ଗୁଣ ସଙ୍ଗେ ସମାନ ହେବ ।
ଅର୍ଥାତ୍ y + 7 = 2x = 2x – y = 7 …(ii)

‘x’ ର, ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, 3 × 12 – 2y = 2 ⇒ 36 – 2y = 2
⇒ 2y = 34 ⇒ y = 17
∴ ପ୍ରଥମ ସଂଖ୍ୟାଟି 12 ଏବଂ ଦ୍ବିତୀୟ ସଂଖ୍ୟାଟି 17 ।

Question 9.
ଗୋଟିଏ ଭଗ୍ନାଂଶର ଲବ ଓ ହରରେ 2 ଯୋଗକଲେ ତାହା \(\frac{9}{11}\) ହୁଏ । ମାତ୍ର ଲବ ଓ ହରରେ 3 ଯୋଗକଲେ \(\frac{5}{6}\)
ହୁଏ । ତେବେ ଭଗ୍ନାଂଶଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାର ଲବ x ଓ ହର y
ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}\)
ପ୍ରଶ୍ବାନୁସାରେ, \(\frac{x+2}{y+2}=\frac{9}{11}\)
⇒ 11x + 22 = 9y + 18 ⇒ 11x – 9y = – 4 (i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x+3}{y+3}=\frac{5}{6}\)
⇒ 6x + 18 = 5y + 15 ⇒ 6x – 5y = – 3 … (ii)

x ର ମାନ ସମୀକରଣ (i)ରେ ସଂସ୍ଥାପନ କଲେ, 11 × 7 – 9y = -4
⇒77 – 9y = 4 ⇒ – 9y = – 4 – 77 = – 81 ⇒ y =\(\frac{81}{9}=9\)
∴ ଭଗ୍ନ ସଂଖ୍ୟାଟି \(\frac{x}{y}=\frac{7}{9}\)

Question 10.
ଗୋଟିଏ ଭଗ୍ନାଂଶର ଲବର 3 ଗୁଣ ଓ ହରରୁ 3 ବିୟୋଗ କଲେ ଭଗ୍ନାଂଶଟି \(\frac{18}{11}\) ହୁଏ । ମାତ୍ର ଲବରେ 8 ଯୋଗକଲେ ଓ ହରକୁ 2 ଗୁଣ କଲେ ତାହା \(\frac{2}{5}\) ହୁଏ । ତେବେ ଭଗ୍ନାଂଶ କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାର ଲବ x ଓ ହର y
∴ ଭଗ୍ନ ସଂଖ୍ୟାଟି \(\frac{x}{y}\)
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{3x}{y-3}=\frac{18}{11}\)
⇒ 33x= 18y – 54 ⇒ 33x – 18y = – 54 ……..(i)
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{x+8}{2y}=\frac{2}{5}\)
⇒ 5x + 40 = 4y ⇒ 5x – 4y = -40 ……….(ii)

x ର ମାନ ସମୀକରଣ (i)ରେ ବସାଇଲେ,
⇒ 33 × 12 – 18y = 54 ⇒ -18y = -54 – 396
⇒ 18y = 450 ⇒ y = \(\frac{450}{18}=25\)
∴ ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}=\frac{12}{25}\)

Question 11.
5ଟି କଲମ ଓ ଟି ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ ମିଶି ୨ ଟଙ୍କା ଏବଂ 3ଟି କଲମ ଓ 2ଟି ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ ମିଶି 5 ଟଙ୍କା ହୁଏ । ତେବେ ଗୋଟିଏ କଲମ ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ କେତେ ?
ସମାଧାନ :
ମନେକର ଗୋଟିଏ କଲମର ଦାମ୍ x ଟଙ୍କା ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ y ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ, 5x + 6y = 9 …… (i)
3x + 2y = 5 …….. (ii)

xର ମାନ ସମୀକରଣ (ii)ରେ ବସାଇଲେ, 3 × \(\frac{3}{2}\) + 2y = 5
⇒ 2y = 5 – \(\frac{9}{2}\) ⇒ 2y = \(\frac{10-9}{2}\)
⇒ y = \(\frac{1}{4}\) ଟଙ୍କା ।
∴ ଗୋଟିଏ କଲମର ଦାମ୍ \(\frac{3}{2}\) ଟଙ୍କା ଓ ଗୋଟିଏ ପେନ୍‌ସିଲ୍‌ର ଦାମ୍ \(\frac{1}{4}\) ଟଙ୍କା ।

Question 12.
ପିତାଙ୍କ ବୟସ ପୁତ୍ର ବୟସର 3 ଗୁଣ । 12 ବର୍ଷ ପରେ ପିତାଙ୍କ ବୟସ ପୁତ୍ର ବୟସର 2 ଗୁଣ ହେବ । ତେବେ ପିତା ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ କେତେ ?
ସମାଧାନ :
ମନେକର ପିତାଙ୍କର ବୟସ x ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ y ବର୍ଷ |
ପ୍ରଶ୍ନନୁସାରେ, x = 3y … (i)
ପୁନଶ୍ଚ 12 ବର୍ଷ ପରେ ପିତାଙ୍କର ବୟସ (x + 12) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ (y + 12) ବର୍ଷ |
ପ୍ରଶ୍ନନୁସାରେ x + 12 = 2 (y + 12) ……..(ii)
ସମୀକରଣ (i)ର ମାନ ସମୀକରଣ (ii)ରେ ସଂସ୍ଥାପନ କଲେ,
3y +12 = 2 (y + 12) ⇒ 3y + 12 = 2y + 24
⇒3y – 2y = 24 – 12
⇒ y = 12 ବର୍ଷ |
ସମୀକରଣ (i)ରେ y = 12 ପ୍ରୟୋଗ କଲେ
x = 3y = 3 × 12 = 36 ବର୍ଷ |
∴ ପିତାଙ୍କର ବର୍ତ୍ତମାନ ବୟସ 36 ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ 12 ବର୍ଷ ।

Question 13.
ଏକ ଆୟତ କ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟକୁ 5 ସେ.ମି. କମାଇ ପ୍ରସ୍ଥକୁ 3 ସେ.ମି. ବଢ଼ାଇବା ଦ୍ଵାରା ଏହାର କ୍ଷେତ୍ରଫଳ 9 ବର୍ଗ ସେ.ମି. କମିଯାଏ । ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟକୁ 3 ସେ.ମି. ଓ ପ୍ରସ୍ଥକୁ 2 ସେ.ମି. ବଢ଼ାଇବା ଦ୍ୱାରା କ୍ଷେତ୍ରଫଳ 67 ବର୍ଗ ସେ.ମି. ବଢ଼ିଯାଏ । ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ x ସେ.ମି. ଓ ପ୍ରସ୍ଥ y ସେ.ମି. ।
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = xy ବର୍ଗ ସେ.ମି.
ପ୍ରଶ୍ନନୁସାରେ, (x – 5) (y + 3) = xy – 9 ⇒ 3x – 5y – 6 = 0 … (i)
ପୁନଶ୍ଚ (x + 3) (y + 2) = xy + 67 = 2x +3y – 61 = 0 …….(ii)

⇒ \(\frac{x}{323}=\frac{y}{171}=\frac{1}{19}\)
⇒ x = \(\frac{323}{19}\) = 17 ଏବଂ y = \(\frac{171}{19}\) = 9
∴ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ 17 ସେ.ମି. ଓ ପ୍ରସ୍ଥ 9 ସେ.ମି. ।
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = ଦୈର୍ଘ୍ୟ × ପ୍ରସ୍ଥ = 17 × 9 = 153 ବର୍ଗ ସେ.ମି. ।

Question 14.
2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ ଲୋକ ଏକତ୍ର ଗୋଟିଏ କାର୍ଯ୍ୟକୁ 5 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ସେହି କାର୍ଯ୍ୟକୁ 4 ଜଣ ପୁରୁଷ ଓ ୨ ଜଣ ସ୍ତ୍ରୀ ଲୋକ ଏକତ୍ର 2 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ତେବେ ଜଣେ ସ୍ତ୍ରୀ ଲୋକ କିମ୍ବା ଜଣେ ପୁରୁଷ ସେହି କାର୍ଯ୍ୟକୁ କେତେ ଦିନରେ ଶେଷ କରିପାରବେ ?
ସମାଧାନ :
ମନେକର ଜଣେ ପୁରୁଷ ଏବଂ ଜଣେ ସ୍ତ୍ରୀ ଗୋଟିଏ କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ x ଓ y ଦିନରେ ଶେଷ କରିପାରିବେ ।
∴ ଜଣେ ପୁରୁଷ ଏବଂ ଜଣେ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଅଂଶ ଏବଂ \(\frac{1}{y}\) ଅଂଶ କରିବେ ।
2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର = \(\frac{2}{x}\) + \(\frac{3}{y}\) ଅଂଶ କରିବେ ।
କିନ୍ତୁ 2 ଜଣ ପୁରୁଷ ଓ 3 ଜଣ ସ୍ତ୍ରୀ କାର୍ଯ୍ୟଟିକୁ 5 ଦିନରେ କରନ୍ତି ।
1 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବେ = \(\frac{1}{5}\) ଅଂଶ ।
ପ୍ରଶାନୁସାରେ, \(\frac{2}{x}+\frac{3}{y}=\frac{1}{5}\) ……….(i)
ପୁନଶ୍ଚ, 4 ଜଣ ପୁରୁଷ ଓ 9 ଜଣ ସ୍ତ୍ରୀ 1 ଦିନରେ କାର୍ଯ୍ୟର କରିବେ = \(\frac{4}{x}\) + \(\frac{9}{y}\) ଅଂଶ ।
କିନ୍ତୁ 4 ଜଣ ପୁରୁଷ ଓ 9 ଜଣ ସ୍ତ୍ରୀ କାର୍ଯ୍ୟଟିକୁ 2 ଦିନରେ କରନ୍ତି ।
1 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବେ = \(\frac{1}{2}\) ଅଂଶ ।
ପୁନଶ୍ଚ, ପ୍ରଶ୍ନନୁସାରେ, \(\frac{4}{x}+\frac{9}{y}=\frac{1}{2}\)

⇒ \(– \frac{3}{y}=\frac{4-5}{10}=\frac{-1}{10}\) ⇒ \(\frac{3}{y}=\frac{1}{10}\)
⇒ y = 30
y ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
⇒ \(\frac{2}{x}+\frac{3}{30}=\frac{1}{5}\) ⇒ \(\frac{2}{x}+\frac{1}{10}=\frac{1}{5}\)
\(\frac{2}{x}+\frac{1}{5}-\frac{1}{10}\) ⇒ \(\frac{2}{x}=\frac{1}{10}\) ⇒ x = 20
∴ ଜଣେ ପୁରୁଷ କିମ୍ବା ଜଣେ ସ୍ତ୍ରୀ ସେହି କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ 20 ଦିନରେ କିମ୍ବା 30 ଦିନରେ କରିବେ ।

Question 15.
A ଓ B ଏକତ୍ର କାମ କରି ଗୋଟିଏ କାର୍ଯ୍ୟକୁ 8 ଦିନରେ ଶେଷ କରିପାରନ୍ତି । ସେମାନେ ଏକତ୍ର କାର୍ଯ୍ୟ ଆରମ୍ଭ କରି 3 ଦିନ କାର୍ଯ୍ୟ କରିବା ପରେ A ଚାଲିଗଲା ଓ ଅବଶିଷ୍ଟ କାର୍ଯ୍ୟକୁ B ଏକା ଆଉ 15 ଦିନରେ ଶେଷ କଲା । ପ୍ରତ୍ୟେକ ଏକାକୀ କାମ କଲେ କେତେ ଦିନରେ କାର୍ଯ୍ୟକୁ ଶେଷ କରିପାରିବେ ?
ସମାଧାନ :
ମନେକର A ଓ B କାର୍ଯ୍ୟକୁ ଯଥାକ୍ରମେ x ଓ y ଦିନରେ କରିପାରିବେ ।
A ଓ B 1 ଦିନରେ କାର୍ଯ୍ୟଟିର ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଅଂଶ ଏବଂ \(\frac{1}{y}\) ଅଂଶ କରିପାରିବେ ।
A ଓ B ମିଶି 1 ଦିନରେ କାର୍ଯ୍ୟଟିର \(\frac{1}{x}+\frac{1}{y}\) ଅଂଶ କରିପାରିବେ ।
କିନ୍ତୁ ପ୍ରଶ୍ନ ଅଛି A ଓ B କାର୍ଯ୍ୟଟିକୁ 8 ଦିନରେ କରିପାରନ୍ତି ।
∴ 1 ଦିନରେ କରିବେ = \(\frac{1}{8}\) ଅଂଶ ।
ପ୍ରଶ୍ନନୁସାରେ, \(\frac{1}{x}+\frac{1}{y}=\frac{1}{8}\) …….. (i)
A ଓ B ଏକତ୍ର କାର୍ଯ୍ୟ ଆରମ୍ଭ କରିବାର 3 ଦିନ ପରେ A ଚାଲିଗଲା ।
ଅବଶିଷ୍ଟ କାର୍ଯ୍ୟଟିକୁ B ଆଉ 15 ଦିନରେ ଶେଷକଲା ।
ଅର୍ଥାତ୍ A, 3 ଦିନ ଓ B (15 + 3) = 18 ଦିନ କାର୍ଯ୍ୟ କଲାପରେ କାର୍ଯ୍ୟଟି ସମ୍ପୂର୍ଣ ହେଲା ।
∴ A, 3 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବ = \(\frac{3}{x}\) ଅଂଶ । B, 18 ଦିନରେ କାର୍ଯ୍ୟଟିର କରିବ \(\frac{18}{y}\) ଅଂଶ ।
ପୁନଶ୍ଚ ପ୍ରଶ୍ନନୁସାରେ, \(\frac{3}{x}+\frac{18}{y}=1\) ……….(ii)

⇒ \(– \frac{15}{y}=\frac{3-8}{8}\) ⇒ \(– \frac{15}{y}=\frac{-5}{8}\)
⇒ \(\frac{3}{y}=\frac{1}{8}\) ⇒ y = 24
y ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ,
⇒ \(\frac{1}{x}+\frac{1}{24}=\frac{1}{8}\) ⇒ \(\frac{1}{x}=\frac{1}{8}-\frac{1}{24}\)
\(\frac{1}{x}=\frac{3-1}{24}=\frac{2}{24}=\frac{1}{12}\) ⇒ x = 12
∴ A ଏକାକୀ କାର୍ଯ୍ୟଟିକୁ 12 ଦିନରେ ଓ B ଏକାକୀ କାର୍ଯ୍ୟଟିକୁ 24 ଦିନରେ ଶେଷ କରିପାରିବେ ।

Question 16.
A ଓ Bର ଆୟର ଅନୁପାତ 8 : 7 ଓ ବ୍ୟୟର ଅନୁପାତ 19 : 16 । ଯଦି ଉଭୟେ 1250 ଟଙ୍କା ସଞ୍ଚୟ କରିପାରନ୍ତି, ତେବେ ସେମାନଙ୍କର ଆୟ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ :
A ଓ B ର ଆୟର ଅନୁପାତ 8 : 7 ।
ମନେକର A ଓ B ର ଆୟ ଯଥାକ୍ରମେ 8x ଟଙ୍କା ଏବଂ 7x ଟଙ୍କା ।
ସେହିପରି A ଓ B ର ବ୍ୟୟର ଅନୁପାତ 19 : 16 |
ତେଣୁ ମନେକର A ଓ B ର ବ୍ୟୟ ଯଥାକ୍ରମେ 19y ଟଙ୍କା ଏବଂ 16y ଟଙ୍କା ।
ପ୍ରଶ୍ନନୁସାରେ A ର ସଞ୍ଚୟ = 1250 ଟଙ୍କା ⇒ 8x – 19y = 1250 ……..(i)
ସେହିପରି B ର ସଞ୍ଚୟ = 1250 ଟଙ୍କା ⇒ 7x – 16y = 1250 ……….(ii)

∴ A = 8x = 8 × 750 = 6000 ଟଙ୍କା । ଏବଂ Bର ଆୟ = 7x = 7 × 750 = 5250 ଟଙ୍କା ।

Question 17.
5 ବର୍ଷ ପରେ ପିତାର ବୟସ ପୁତ୍ରର ବୟସର ତିନିଗୁଣ ହେବ ଓ 5 ବର୍ଷ ପୂର୍ବେ ପିତାର ବୟସ ପୁତ୍ର ବୟସର ସାତଗୁଣ ଥିଲା । ତେବେ ସେମାନଙ୍କର ବର୍ତ୍ତମାନ ବୟସ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ପିତାର ବର୍ତ୍ତମାନ ବୟସ x ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ y ବର୍ଷ ।
5 ବର୍ଷ ପରେ ପିତାର ବୟସ ହେବ = (x + 5) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ ହେବ = (y + 5) ବର୍ଷ
ପ୍ରଶ୍ନନୁସାରେ, x + 5 = 3(y + 5)
⇒ x + 5 = 3y + 15 ⇒ x – 3y = 10 …(i)
ପୁନଶ୍ଚ, 5 ବର୍ଷ ପୂର୍ବେ ପିତାର ବୟସ ଥିଲା = (x – 5) ବର୍ଷ ଓ ପୁତ୍ରର ବୟସ ଥିଲା = (y – 5) ବର୍ଷ
ପ୍ରଶ୍ନନୁସାରେ, (x – 5) = 7(y – 5)
⇒ x -5 = 7y – 35 ⇒ x – 7y = -30 …(ii)
ସମୀକରଣ (i)ରୁ ସମୀକରଣ (ii)କୁ ବିୟୋଗ କଲେ, (x – 3y) – (x – 7y) = 10 – (-30)
⇒ x – 3y – x + 7y = 40 ⇒ 4y = 40 ⇒ y = 10
ସମୀକରଣ (i) ରେ y = 10 ସ୍ଥାପନ କଲେ, x – 3 × 10 = 10 ⇒ x = 40
∴ ପିତାର ବର୍ତ୍ତମାନ ବୟସ 40 ବର୍ଷ ଓ ପୁତ୍ରର ବର୍ତ୍ତମାନ ବୟସ 10 ବର୍ଷ ।

Question 18.
ଗୋଟିଏ ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ 2 ମି. ଅଧ୍ଵ ଓ ପ୍ରସ୍ଥ 2ମି. କମ୍ ହେଲେ, କ୍ଷେତ୍ରଫଳ 28 ବ.ମି. କମିଯାଏ; ମାତ୍ର ଦୈର୍ଘ୍ୟ 1 ମି. କମ୍ ଓ ପ୍ରସ୍ଥ 2 ମି. ଅଧ୍ବକ ହେଲେ କ୍ଷେତ୍ରଫଳ 33 ବ. ମି. ବଢ଼ିଯାଏ । ମୂଳ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ ସ୍ଥିର କର ।
ସମାଧାନ :
ମନେକର ଆୟତକ୍ଷେତ୍ରର ଦୈର୍ଘ୍ୟ ଓ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ x ମି. ଓ Y ମି. ।
∴ କ୍ଷେତ୍ରଫଳ = ଦୈର୍ଘ୍ୟ × ପ୍ରସ୍ଥ = xy ବର୍ଗ ମି. ।
ପ୍ରଶ୍ନନୁସାରେ, ଦୈର୍ଘ୍ୟ 2 ମି. ଅଧୂକ ଏବଂ ପ୍ରସ୍ଥ 2 ମି. କମ୍ ହେଲେ କ୍ଷେତ୍ରଫଳ 28 ବର୍ଗ ମି. କମିଯାଏ ।
ତେଣୁ ପରିବର୍ତ୍ତିତ ଦୈର୍ଘ୍ୟ ଏବଂ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ (x + 2) ମି. ଏବଂ (y – 2) ମି. ହେବ ।
(x + 2) (y – 2) = xy – 28 ⇒ xy + 2y – 2x – 4 = xy – 28
⇒ 2y – 2x = -28 + 4 ⇒ y – x = \(\frac{-24}{2}\) ⇒ y – x = -12
⇒ x – y = 12 … (i)
ସେହିପରି ଦୈର୍ଘ୍ୟ 1 ମି. କମ୍ ଓ ପ୍ରସ୍ଥ 2 ମି. ଅଧ୍ଯକ ହେଲେ, କ୍ଷେତ୍ରଫଳ 33 ବର୍ଗ ମି. ବୃଦ୍ଧିପାଏ ।
∴ ପରିବର୍ତ୍ତିତ ଦୈର୍ଘ୍ୟ ଏବଂ ପ୍ରସ୍ଥ ଯଥାକ୍ରମେ (x – 1) ମି, ଏବଂ (y + 2) ମି. ହେବ ।
∴ (x – 1) (y + 2) = xy + 33 ⇒ xy + 2x – y – 2 = xy + 33
⇒ 2x – y = 33 +2 ⇒ 2x – y = 35 ……(ii)

‘x’ ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ 23 – y = 12 = y = 11
∴ ଆୟତକ୍ଷେତ୍ରର କ୍ଷେତ୍ରଫଳ = xy ବର୍ଗ ମି. = (23 × 11) ବର୍ଗ ମି. = 253 ବର୍ଗ ମି. ।

Question 19.
50କୁ ଏପରି ଦୁଇଟି ସଂଖ୍ୟାର ସମଷ୍ଟି ରୂପେ ପ୍ରକାଶ କର ଯେପରିକି ସଂଖ୍ୟା ଦ୍ଵୟର ବ୍ୟକ୍ରମର ସମଷ୍ଟି \(\frac{1}{12}\) ହେବ ।
ସମାଧାନ :
ମନେକର ସଂଖ୍ୟା ଦୁଇଟି x ଓ y ।
ସଂଖ୍ୟା ଦୁଇଟିର ସମଷ୍ଟି 50 । ⇒ x + y = 50 ….. (i)
x ଓ y ର ବ୍ୟକ୍ରମ ଯଥାକ୍ରମେ \(\frac{1}{x}\) ଓ \(\frac{1}{y}\) ।
ପ୍ରଶାନୁସାରେ, \(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ……..(ii)
⇒ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ⇒ \(\frac{x+y}{xy}=\frac{1}{12}\) ⇒ \(\frac{50}{xy}=\frac{1}{12}\) (∵ x + y = 50)
⇒ xy = 600 ……..(iii)
ଆମେ ଜାଣିଛେ, x – y = \(\sqrt{(x+y)^2-4xy}\) = \(\sqrt{(50)^2-4×600}\) = \(\sqrt{100}\)
⇒ x – y = 10 ……(iv)
ସମୀକରଣ (i) ଓ (iv)କୁ ଯୋଗକଲେ, 2x = 60 ⇒ x = 30
ସମୀକରଣ (i)ରେ x = 30 ସ୍ଥାପନ କଲେ, 30 + y = 50 ⇒ y = 20
∴ ସଂଖ୍ୟାଦ୍ବୟ 30 ଓ 20 ।

Question 20.
ଗୋଟିଏ ଭଗ୍ନ ସଂଖ୍ୟାର ଲବ ଓ ହରକୁ ଯୋଗକରି ଯୋଗଫଳର ଏକ-ତୃତୀୟାଂଶ ନେଲେ, ତାହା ହରଠାରୁ 4 ଊଣା ହୁଏ ଓ ହରରେ 1 ଯୋଗକରି ଭଗ୍ନ ସଂଖ୍ୟାଟିକୁ ଲଘିଷ୍ଠ ଆକାରରେ ଲେଖୁଲେ ତାହା \(\frac{1}{4}\) ହୁଏ । ଭଗ୍ନ ସଂଖ୍ୟାଟି କେତେ ?
ସମାଧାନ :
ମନେକର ଭଗ୍ନସଂଖ୍ୟାଟି \(\frac{x}{y}\), ଯାହାର ଲବ x ଏବଂ ହର y ।
ପ୍ରଶ୍ନନୁସାରେ, ଲବ ଓ ହରକୁ ଯୋଗକରି ଯୋଗଫଳର ଏକତୃତୀୟାଂଶ ନେଲେ ତାହା ହରଠାରୁ 4 ଊଣା ହୁଏ ।
\(\frac{1}{3}\)(x+y) = y – 4 ⇒ x + y = 3y – 12 ⇒ x – 2y= – 12 …….. (i)
ପୁନଶ୍ଚ ହରରେ 1 ଯୋଗକରି ଭଗ୍ନାଂଶଟିକୁ ଲଘିଷ୍ଠ ଆକାରକୁ ଆଣିଲେ ତାହା \(\frac{1}{4}\) ହୁଏ ।
\(\frac{x}{y+1}=\frac{1}{4}\) ⇒ 4x = y + 1 ⇒ 4x – y = 1 ……….(ii)

‘y’ ର ମାନ ସମୀକରଣ (i) ରେ ପ୍ରୟୋଗ କଲେ, x – 2 × 7 = -12
⇒ x – 14 = -12 ⇒ x = 2
∴ ଭଗ୍ନାଂଶଟି \(\frac{x}{y}=\frac{2}{7}\)

Odisha State Board BSE Odisha 6th Class English Solutions Test-1(B) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1(B)

BSE Odisha 6th Class English Test-1(B) Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.
1. Write the following Odia names of the persons in English. (Teacher will give the names of six persons in Odia.) [06]
Answer:
ଲାଲା ଲଜପାଟ ରୟ – Lala Lajpat Roy
B.R. ଆମ୍ବେଦକର | – B.R. Ambedkar
ବାଙ୍କିମ୍ ଚନ୍ଦ୍ର ପାଲ୍ – Bankim Chandra Pal.
ବୁକ୍ସି ଜଗବନ୍ଧୁ – Buxi Jagabandhu
ଫାକିର ମୋହନ ସେନାପତି – Fakir Mohan Senapati
ରାଧନାଥା ରାଧା – Radhanatha Ratha

2. Write the following place names in English. [06]
(Teacher will give names of six places in Odia.)
Answer:
ଶ୍ରୀନଗର – Srinagar
ହିମାଚଳ ପ୍ରଦେଶ – Himachal Pradesh
ପୁଡୁଚେରୀ – Puducheri
ଦ୍ୱାରିକା – Dwarika
ଚିଲିକା ହ୍ରଦ – Chilika lake
ନନ୍ଦନକାନନ୍ ପ୍ରାଣୀ ଉଦ୍ୟାନ – Nandankanan Zoo

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]
Answer :
Jagannatha     Moon
Saraswati        Capital
Ganesha         Chief-Minister
Earth /World  President
Sun                 Garden

4. Given below are some words. Your teacher will read aloud five of them.
Tick those which s/he reads aloud. [05]
temper, catch, squirrel, little, agile, climbed, thrash, burst, bamboo, curry, foolish.
[Listen to your teacher and tick those words s/he reads.]

5. Your teacher will read aloud a paragraph. You listen to him/her and fill in the gaps. (Question with Answer) [08]
Once there lived a greedy fat old man. One day he got up at 6 a.m. and brushed his teeth at 6.30 a.m. He took tea at 7 a.m. and breakfast at 8.30 a.m. Do you know how much tea he took?

6. Match the words which sound alike at the end. (Question with Answer)

7. Read the poem and answer the questions. [10]
Their ears are pink,
Their teeth are white,
They run about
The house at night
They nibble things
They shouldn’t touch
And no one seems
To like them much.

Question (i).
What is the color of their ears?
Answer:
The color of their ears is pink.

Question (ii).
What is the color of their teeth?
Answer:
The color of their teeth is white.

Question (iii).
Where do they run about at night?
Answer:
They run about the house at night.

Question (iv).
Which things do they nibble?
Answer:
They nibble things that they shouldn’t touch.

Question (v).
Do most people like mice?
Answer:
No, no one seems to like mice.

8. Read the following paragraph and answer the questions in complete sentences. [20]
After eating the dog, the man walked, walked, and walked till he met a little squirrel. The little squirrel asked the old man, “Old man, old man, what makes you so fat? The old man said, “I’ve taken a very heavy breakfast. In my breakfast, I took two mugs of tea, two liters of milk, three tins of biscuits, and five big pieces of cakes.” Then I ate a little boy and a small dog. I’ll also eat you up if I can catch you”.

“But you cannot catch me, old man,” said the active, agile, little squirrel. Then the squirrel jumped up the tree, the old man also climbed up the tree. The little squirrel jumped up to the main branch of the tree. The old man also climbed up to the main branch of the tree. Next, the little squirrel jumped up to a thin branch. The old man also climbed up to the thin branch. But thrash ! the small branch broke and the old man fell to the ground. His big belly burst out.

Question (i).
Whom did the old man meet in this section?
Answer:
In this section, the old man met a little squirrel.

Question (ii).
What did the squirrel ask the old man?
Answer:
The squirrel asked the old man, “What makes you so fat?”

Question (iii).
How many mugs of tea did the old man take?
Answer:
The old man took two mugs of tea.

Question (iv).
How much milk did he take?
Answer:
He took two liters of milk.

Question (v).
How many tins of biscuits did he take in his breakfast?
Answer:
For his breakfast, he took three tins of biscuits.

Question (vi).
How many pieces of cake did he take?
Answer:
He took five big pieces of cake.

Question (vii).
Where did the squirrel jump up first?
Answer:
The squirrel jumped up first to the tree.

Question (viii).
What did the old man do?
Answer:
The old man also climbed up the tree.

Question (ix).
Where did the squirrel jump up then?
Answer:
Then the squirrel jumped up to the main branch of the tree.

Question (x).
What happened to the old man when he fell down?
Answer:
When the old man fell down, his big belly burst out.

9. Read the following poem and answer the questions in complete sentences. [10]
“My story is said
The flowering plant is dead.
O flowering plant! Why did you die?
The black cow ate me up and made me lie.
O black cow! Why did the plant you eat?
Because the cowherd did not do well treat.
O, cowherd! Why didn’t you well the cow treat to eat?
The daughter-in-law did not give me food.”

Question (i).
Why is the flowering plant dead?
Answer:
The flowering plant is dead because the black cow ate it up.

Question (ii).
Why did the cow eat up the flowering plant?
Answer:
The cow ate up the flowering plant because the cowherd did not well her treat.

Question (iii).
Why didn’t the cowherd well the cow treat to eat?
Answer:
The cowherd didn’t well the cow treat to eat because the daughter-in-law did not give him food.

Question (iv).
What is said?
Answer:
Your story is said.

Question (v).
What type of plant is dead?
Answer:
The flowering plant is dead.

10. Read the following paragraph and answer the questions in a complete sentences. [20]
Once a foolish Santal son-in-law went to his in-law’s house. His mother-in-law cooked delicious dishes for her son-in-law. One of the dishes was a curry made out of the bamboo shoot. The son-in-law liked it very much and asked his mother-in-law, “Mother, the curry is extremely delicious. What is the curry made from? Instead of answering his question, she pointed at the bamboo door. The son-in-law asked, “Is it from bamboo ?” Yes, son, the curry is made from bamboo and is, therefore, called “Bamboo Curry”. The next day, the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo. So he carried home the bamboo door of his in-law’s house.

Question (i).
Who went to his father-in-law’s house?
Answer:
A foolish Santal son-in-law went to his father-in-law’s house.

Question (ii).
What did his mother-in-law cook for him?
Answer:
His mother-in-law cooked delicious dishes for him.

Question (iii).
What was one of the dishes made?
Answer:
One of the dishes was a curry made out of the bamboo shoot.

Question (iv).
How was the curry?
Answer:
The curry was extremely delicious.

Question (v).
What did the son-in-law ask his mother-in-law?
Answer:
The son-in-law asked his mother-in-law what the curry was made from.

Question (vi).
What did the mother-in-law answer?
Answer:
Instead of answering his question, she pointed at the bamboo door.

Question (vii).
Why is the curry called “Bamboo Curry”?
Answer:
The curry is called ‘Bamboo Curry’ because it is made from bamboo.

Question (viii).
When was the son-in-law returning to his home?
Answer:
The next day, the son-in-law was returning to his home.

Question (ix).
What did he think of doing?
Answer:
He thought of cooking bamboo curry at home.

Question (x).
What did not they have? What did he do for that?
Answer:
They did not have bamboo. So he carried home the bamboo door of his in-law’s house.

Odisha State Board BSE Odisha 6th Class English Solutions Test-1(A) Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1(A)

BSE Odisha 6th Class English Test-1(A) Text Book Questions and Answers

The figures in the right-hand margin indicate the marks for each question.

1. Write the following Odia names of the persons in English. [06]
(Teacher will give the names of six persons in Odia.)
Answer:
ଲାଲ ବାହାଦୂର ଶାସ୍ତ୍ରୀ | – Lal Bahadur Shastri
ବାଲ ଗଙ୍ଗାଧର ତିଲକ | – Bal Gangadhar Tilak
ସର୍ଦ୍ଦାର ଭାଲାଭଭାଇ ପଟେଲ | – Sardar Ballavabhai Patel
ହରେକୃଷ୍ଣ ମହାତ୍ମା – Harekrushna Mahatab
ଭୀମା ଭୋଇ – Bhima Bhoi
ପ୍ରାଣକୃଷ୍ଣ ପରଜା | – Pranakrushna Parija

2. Write the following place names in English. [06]
(Teacher will give names of six places in Odia.)
Answer:
ହିମାଳୟ – Himalaya
ପଞ୍ଜାବ – Punjab
ଅରୁଣାଚଳ ପ୍ରଦେଶ – Arunachal Pradesh
କୋଲକାତା – Kolkata
ବ୍ରହ୍ମପୁର – Berhampur
ବାଲାସୋର – Balasore

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]
Answer:
Grandfather      Ice-cream       Friend
Grandmother   Computer       Neighbour
Bread               Blackboard      Government
Mango

4. Given below are some words. Your teacher will read aloud five of them.
Tick those which s/he reads aloud. [05]
kite, paper, greedy, breakfast, tea, mugs, biscuits, cakes, conversation, heavey, squirrel.
[Listen to your teacher carefully and tick those words as he reads aloud.]

5. Your teacher will read aloud a paragraph. You listen to him/her and nil in the gaps. (Question with Answer) [08]
Answer:
The old man said, “I’ve taken two mugs of tea, and two liters of milk. I also took three tins of biscuits and five kilograms of cakes. And if I can catch you, I’ll eat you up .’’Then the old man caught the thin little boy and ate him up.

6. Match the words which sound alike at the end. (Question with Answer) [10]

7. Read the poem and answer the questions. [10]
I think mice
Are rather nice.
Their tails are long
Their faces small,
They haven’t any
Skins at all.

Question (i).
What is this poem about?
Answer:
This poem is about mice.

Question (ii).
Who is ‘I’ at the beginning of the poem?
Answer:
‘I’ at the beginning of the poem is the poet himself.

Question (iii).
What are the tails of mice like?
Answer:
The tails of mice are long.

Question (iv).
How are their faces?
Answer:
Their faces are small.

Question (v).
They haven’t any skint at all”- What does this line mean?
Answer:
‘‘They haven’t any skins at all”- This line means that mice have very thin skins.

8. Read the following paragraph and answer the questions in complete sentences. [20]
Once there lived a greedy fat old man. One day he got up at 6 a.m. and brushed his teeth at 6.30 a.m. He took tea at 7 a.m. and breakfast at 8.30 a.m. He took two mugs of tea and two liters of milk. Then he took three tins of biscuits and five big pieces of cake. After breakfast, he looked really very very fat.

Question (i).
What is this paragraph about?
Answer:
This paragraph is about a greedy fat old man.

Question (ii).
How was the old man?
Answer:
The old man was greedy and fat.

Question (iii).
When did he get up one day?
Answer:
One day he got up at 6 a.m.

Question (iv).
When did he brush his teeth?
Answer:
He brushed his teeth at 6.30 a.m.

Question (v).
What did he take at 7 a.m.?
Answer:
He took tea at 7 a.m.

Question (vi).
When did he take his breakfast?
Answer:
He took his breakfast at 8.30 a.m.

Question (vii).
How many mugs of tea did he take?
Answer:
He took two mugs of tea.

Question (viii).
How much milk did he take?
Answer:
He took two liters of milk.

Question (ix).
How many tins of biscuit did he take?
Answer:
He took three tins of biscuits.

Question (x).
How did he look like after breakfast?
Answer:
After breakfast, he looked really very very fat.

9. Read the following poem and answer the questions in complete sentences. [10]

For want of a nail,
the shoe was lost.
For want of a shoe,
the horse was lost.
For want of a horse,
the rider was lost.
For want of a rider,
the battle was lost.
For want of a battle,
the kingdom was lost.
And all for the want of a
horseshoe nail.

Question (i).
Why was the shoe lost?
Answer:
The shoe was lost for want of a nail.

Question (ii).
Why was the horse lost?
Answer:
The horse was lost for want of a shoe.

Question (iii).
Why was the rider lost?
Answer:
The rider was lost for want of a horse.

Question (iv).
Why was the battle lost?
Answer:
The battle was lost for want a rider.

Question (v).
Why was the kingdom lost?
Answer:
The kingdom was lost for want of a battle.

10. Read the following paragraph and answer the questions in a complete sentences. [20]
A son-in-law, after his marriage, was planning to visit his father-in-law’s house for the first time. A man from his village gave him the advice, “Use big and high-sounding words in your father-in-law’s house. Always sit on a high place. First say ‘No’ to any food given to you.”
He, therefore, used very long and high-sounding words. He told his mother-in-law, “You are the sweetest, kindest, greatest, and gentlest lady.” The mother-in-law was very much pleased to hear this. She praised her son-in-law in front of her neighbors for using high-sounding words and calling her the kindest and greatest lady.

Question (i).
What was the son-in-law’s plan?
Answer:
The son-in-law’s plan was to visit his father-in-law’s house.

Question (ii).
When was the son-in-law planning to visit his father-in-law’s house?
Answer:
After his marriage, the son-in-law was planning to visit his father-in-law’s house.

Question (iii).
Did he visit his father-in-law’s house before?
Answer:
No, he did not visit his father-in-law’s house before.

Question (iv).
Who advised him to do something at his in-law’s house?
Answer:
A man from his village advised him to do something at his in-law’s house.

Question (v).
What was the first advice?
Answer:
The first piece of advice was to use big and high-sounding words in his father-in-law’s house.

Question (vi).
What was the second advice for the son-in-law?
Answer:
The second piece of advice for the son-in-law was always to sit on a high place.

Question (vii).
What was the third piece of advice for the son-in-law?
Answer:
The third piece of advice for the son-in-law was first to say ‘No’ to any food given to him.

Question (viii).
What did he tell his mother-in-law?
Answer:
He told his mother-in-law that she was the sweetest, kindest, greatest and gentlest lady.

Question (ix).
How did the mother-in-law feel to hear his words?
Answer:
The mother-in-law was very much pleased to hear his words.

Question (x).
What did she do?
Answer:
She praised her son-in-law in front of her neighbors for calling her the kindest and greatest lady.

Odisha State Board BSE Odisha 6th Class English Solutions Test-1 Textbook Exercise Questions and Answers.

BSE Odisha Class 6 English Solutions Test-1

BSE Odisha 6th Class English Test -1 Text Book Questions and Answers

• The figures in the right-hand margin indicate the marks for each question.

1. Write the following Odia names of the persons in English.
(Teacher will give the names of six persons in Odia.) [06]

Answer:
ମହାତ୍ମା ଗାନ୍ଧୀ – Mahatma Gandhi
ଜବାହରଲାଲ ନେହେରୁ | – Jawaharlal Nehru
ସୁଭାଷ ବୋଷ | – Subhas Bose
ଗୋପାବନ୍ଧୁ ଦାସ – Gopabandhu Das
ମଧୁସୂଦନ ଦାସ – Madhusudan Das
ବିଜୁ ପଟ୍ଟନାୟକ – Biju Patnaik

2. Write the following place names in English.
(Teacher will give names of six places in Odia.) [06]

Answer:
ଦିଲ୍ଲୀ – Delhi
କାଶ୍ମୀର – Kashmir
ଅୟୋଧ୍ୟା – Ayodhya
ଆସାମ | – Assam
ଭୁବନେଶ୍ୱର – Bhubaneswar
କଟକ | – Cuttack

3. Your teacher will give a dictation of ten words. Write them in the space given below. [05]

Answer :
School     Mother     Town
Book        House      Temple
Teacher    Village      God
Father

4. Given below are some words. Your teacher will read aloud five of them. Tick those which s/he reads aloud. [05]
house, gentle, great, night, cotton, lemon, nibble, thief, delicious, curry, kingdom, nail, battle, ground.
[Listen to your teacher and tick those words your teacher reads aloud.]

5. Your teacher will read aloud a paragraph. You listen to him/her and fill in the gaps. [08]
(Question with Answer)
Next ____________the son-in-law was ___________to leave for his ______________ . The bamboo _____________came to his ___________________. He thought of_____________ bamboo curry at home. But ________________ did not have _______________________.
Answer:
Next day the son-in-law was about to leave for his home. The bamboo curry came to his mind. He thought of cooking bamboo curry at home. But they did not have bamboo.

6. Match the words which sound alike at the end. (Question with Answer) [10]
(i)

Answer:

(ii)

Answer:

7. Read the poem and answer the questions. [10]

A kite on the ground
Is just paper and string
But up in the air
Will dance and sing.
A kite in the air
Will dance and caper
But back on the ground
Is just string and paper.

Question (a).
What is this poem about?
Answer:
This poem is about a kite.

Question (b).
What does it do when up in the air?
Answer:
When Up in the air, it will dance and sing.

Question (c).
What is it when it is on the ground?
Answer:
When on the ground, it is just paper and string.

Question (d).
Do you like flying a kite?
Answer:
Yes, I like flying a kite very much.

Question (e).
What are kites normally made of?
Answer:
Normally, kites are made of paper and string.

8. Read the following paragraph and answer the questions in complete sentences. [20]
A son-in-law after his marriage was planning to visit his father-in-law for the first time. A man from his village gave him some advice, “Use big and high-sounding words in your father-in-law’s house. Always sit in a high place. First, say ‘no’ to any food given to you.’’

Question (a).
What is the paragraph about?
Answer:
The paragraph is about a son-in-law.

Question (b).
What was he planning?
Answer:
He was planning to visit his father-in-law’s house.

Question (c).
When was he planning?
Answer:
He was planning after his marriage.

Question (d).
Had he gone to his father-in-law’s house before?
Answer:
No, he had not gone to his father-in-law’s house before.

Question (e).
Who gave him some advice?
Answer:
A man from his village gave him some advice.

Question (f).
How many pieces of advice did he give?
Answer:
He gave three pieces of advice.

Question (g).
What was the first advice?
Answer:
The first piece of advice was to use big and high-sounding words in his father-in-law’s house.

Question (h).
What was the second piece of advice?
Answer:
The second piece of advice was always to sit in a high place.

Question (i).
What was the third piece of advice?
Answer:
The third piece of advice was first to say ‘no’ to any food given to him.

Question (j).
Will he carry out the advice?
Answer:
Yes, he will carry out the advice.

9. Read the following poem and answer the questions in complete sentences. [10]

I woke up this morning
And I got out of bed,
Then I took a cup of tea
And ate a slice of bread.
1 went to the bus stop
And caught the bus to school,
On my way back it rained
And the weather was cool.

Question (i).
Who is T in the poem?
Answer:
T in the poem is the poet.

Question (ii).
What did s/he take after getting up?
Answer:
After getting up, s/he took a cup of tea.

Question (iii).
What did s/he eat?
Answer:
S/he ate a slice of bread.

Question (iv).
How did s/he go to school?
Answer:
S/he went to school by bus.

Question (v).
What happened on his / her way back?
Answer:
On his/her way back, it rained and the weather was cool.

10. Read the following paragraph and answer the questions in complete sentences. [20]
There was a crow and there was a cuckoo. They lived together happily. The crow was hard-working. But the cuckoo was very lazy. The crow brought food. The cuckoo only ate. The crow built a nest. She laid her eggs there. The cuckoo also wanted to lay eggs. But she had
not built a nest. One day the crow was not in her nest. The cuckoo threw away the eggs and laid her own eggs there. The crow did not know this. She sat over the eggs for some days. Young ones came out. She took care of them. But when the young ones grew? up, they sang like cuckoos. So she drove them away. From that day the crow always drives away the cuckoo.

Question (i).
What is this paragraph about?
Answer:
This paragraph is about a crow and a cuckoo.

Question (ii).
Who was lazy?
Answer:
The cuckoo was very lazy.

Question (iii).
Who was hard-working?
Answer:
The crow was hard-working.

Question (iv).
Who brought food?
Answer:
The crow brought food.

Question (v).
Who ate it?
Answer:
The cuckoo only ate it.

Question (vi).
Who built a nest?
Answer:
The crow built a nest.

Question (vii).
What did the cuckoo do with the eggs of the crow?
Answer:
The cuckoo threw away the eggs of the crow and laid her own eggs there.

Question (viii).
Whose young ones the crow was taking care of?
Answer:
The crow was taking care of the young ones of the cuckoo.

Question (ix).
When did she come to know about this?
Answer:
When the young ones grew up and sang like cuckoos, she (crow) came to know about this.

Question (x).
Why could not she recognize them before?
Answer:
She (the crow) could not recognize them before because in her absence the cuckoo had thrown away her eggs and laid her (cuckoo’s) own eggs there. Further, the young ones of the crow and the cuckoo look alike.

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 11 Straight Lines Ex 11(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(b)

Question 1.
Fill in the blanks in each of the following, using the answers given against each of them :
(a) The slope and x-intercept of the line 3x – y + k = 0 are equal if k = _________ . (0, -1, 3, -9)
Solution:
-9

(b) The lines 2x – 3y + 1 = 0 and 3x + ky – 1=0 are perpendicular to each other if k = ___________ . (2, 3, -2, -3)
Solution:
2

(c) The lines 3x + ky – 4 = 0 and k – Ay – 3x = 0 are coincident if k = _____________. (1, -4, 4, -1)
Solution:
4

(d) The distance between the lines 3x – 1 = 0 and x + 3 = 0 is _________ units. (4, 2, \(\frac{8}{3}\), \(\frac{10}{3}\))
Solution:
\(\frac{10}{3}\)

(e) The angle between the lines x = 2 and x – √3y + 1 = 0 is _________. (30°, 60°, 120°, 150°)
Solution:
60°

Question 2.
State with reasons which of the following are true or false :
(a) The equation x = k represents a line parallel to x – axis for all real values of k.
Solution:
False. As the line x = k is parallel to y- axis for all values of k.

(b) The line, y + x + 1 = 0 makes an angle 45° with y – axis.
Solution:
y + x + 1 = 0
∴ Its slope = -1 = tan 135°
∴ It makes 45° with y – axis, as it makes 135° with x – axis. (True)

(c) The lines represented by 2x – 3y + 1 = 0 and 3x + 2y – k = 0 are perpendicular to each other for positive values of k only.
Solution:
2x – 3y + 1 = 0, 3x + 2y – k = 0
∴ \(m_1 m_2=\frac{2}{3} \times \frac{(-3)}{2}=-1\)
∴ The lines are perpendicular to each other for + ve values of k only. (False)

(d) The lines represented by px + 2y – 1 = 0 and 3x + py + 1 = 0 are not coincident for any value of ‘p’.
Solution:
px + 2y – 1 = 0, 3x + py + 1=0
∴ \(\frac{p}{3}=\frac{2}{p}=\frac{-1}{1} \Rightarrow p^2=6\)
and p = -3 or -2
There is no particular value of p for which \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) (True)

(e) The equation of the line whose x and y – intercepts are 1 and -1 respectively is x – y + 1 = 0.
Solution:
Equation of the line whose intercepts 1 and -1 is \(\frac{x}{1}+\frac{y}{-1}\) = 1
or, x – y = 1 (False)

(f) The point (-1, 2) lies on the line 2x + 3y – 4 = 0.
Solution:
Putting x = – 1, y = 2
we have 2 (- 1) + 3 × 2 – 4
= -2 + 6 – 4 = 0
∴ The point (-1, 2) lies on the line 2x + 3 – 4 = 0 (True)

(g) The equation of a line through (1, 1) and (-2, -2) is y = – 2x.
Solution:
The equation of the line through (1, 1) and (-2, -2) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
or, y – 1 = \(\frac{-2-1}{-2-1}\) (x – 1)
or, y – 1 = x – 1
or, x – y = 0 (False)

(h) The line through (1, 2) perpendicular to y = x is y + x – 2 =0.
Solution:
The slope of the line y = x is 1.
∴ The slope of the line perpendicular to the above line is -1.
∴ The equation of the line through (1, 2) having slope – 1 is y – y₁ = m(x – x₁)
or, y – 2 = -1 (x – 1)
or, y – 2= -x + 1
or, x + y = 3 (False)

(i) The lines \(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{y}{a}-\frac{x}{b}\) = 1 are intersecting but not perpendicular to each other.
Solution:
\(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{y}{a}-\frac{x}{b}\) = 1
∴ \(m_1 m_2=\frac{\left(-\frac{1}{a}\right)}{\frac{1}{b}} \times \frac{\left(-\frac{1}{b}\right)}{\left(-\frac{1}{a}\right)}=-1\)
∴ The lines intersect and are perpendicular to each other. (False)

(j) The points (1, 2) and (3, – 2) are on the opposite sides of the line 2x + y = 1.
Solution:
2x + y = 1
Putting x = 1, y = 2,
we have 2 × 1 + 2 = 4 > 1
Putting x – 3, y = -2,
we have 2 × 3 – 2 – 4 > 1
∴ Points (1, 2) and (3, – 2) lie on the same side of the line 2x + y = 1 (False)

Question 3.
A point P (x, y) is such that its distance from the fixed point (α, 0) is equal to its distance from the y – axis. Prove that the equation of the locus is given by, y² = α (2x – α).
Solution:

Question 4.
Find the locus of the point P (x, y) such that the area of the triangle PAB is 5, where A is the point (1, -1) and B is the tie point (5, 2).
Solution:

= \(\frac{1}{2}\) (-3x + 4y + 7) = 5
or, – 3x + 4y + 7 = 10
or, 3x – 4y + 3 =0 which is the locus of the point P (x, y).

Question 5.
A point is such that its distance from the point (3, 0) is twice its distance from the point (-3, 0). Find the equation of the locus.

Question 6.
Obtain the equation of straight lines:
(a) Passing through (1, – 1) and making an angle 150°.
Solution:
The slope of the line
= tan 150° = –\(\frac{1}{\sqrt{3}}\)
∴ The equation of the line is y – y₁ = m(x – x₁)
or, y + 1 = –\(\frac{1}{\sqrt{3}}\) (x – 1)
or, y√3 + √3 = -x + 1
or, x + y√3 + √3 – 1 = 0

(b) Passing through (-1, 2) and making intercept 2 on the y-axis.
Solution:
Let the equation of the line be
y – mx + c or, y = mx + 2
∴ As the line passes through (-1, 2)
we have 2 = – m + 2, or, m = 0
∴ Equation of the line is y = 2.

(c) Passing through the points (2, 3) and (-4, 1).
Solution:
The equation of the line is

(d) Passing through (- 2, 3) and a sum of whose intercepts in 2.
Solution:
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}\) = 1 where a + b = 2     …….(1)
Again, as the line passes through the point (-2, 3), we have \(\frac{-2}{a}+\frac{3}{b}\) = 1      ………(2)
From (1), we have a= 2 – b
∴ From (2) \(\rightarrow \frac{-2}{2-b}+\frac{3}{b}=1\)
or, – 2b + 6 – 3b = (2 – b)b
or, 6 – 5b = 2b – b²
or, b² – 7b + 6 = 0
or, (b – 6)(b – 1) = 0
∴ b = 6, 1
∴ a =2 – b = 2 – 6 = -4
or, 2 – 1 = 1
∴ Equation of the lines are \(\frac{x}{-4}+\frac{y}{6}\) = 1 or \(\frac{x}{1}+\frac{y}{1}\) = 1
i, e. -3x + 2y = 12 or, x + y = 1

(e) Whose perpendicular distance from the origin is 2 such that the perpendicular from the origin has indication 150°.
Solution:
Here p = 2, α = 150°
The equation of the line in normal form is x cos α + y sin α = p
or, x cos 150° + y sin 150° = 2
or, \(\frac{-x \sqrt{3}}{2}+y \cdot \frac{1}{2}\) = 2
or, -x √3 + y = 4
or, x√3 – y + 4 = 0

(f) Bisecting the line segment joining (3, – 4) and (1, 2) at right angles.
Solution:
The slope of the line \(\overline{\mathrm{AB}}\) is

(g) Bisecting the line segment joining, (a, 0) and (0, b) at right angles.
Solution:
Refer to (f)

(h) Bisecting the line segments joining (a, b), (a’, b’) and (-a, b), (a’, -b’).
Solution:

(i) Passing through the origin and the points of trisection of the portion of the line 3x + y – 12 = 0 intercepted between the coordinate axes.
Solution:

(j) Passing through (-4, 2) and parallel to the line 4x – 3y = 10.
Solution:
Slope of the line 4x – 3y = 10 is \(\frac{-4}{-3}=\frac{4}{3}\)
∴ The slope of the line parallel to the above line is \(\frac{4}{3}\).
∴ Equation of the line through (- 4, 2) and having slope \(\frac{4}{3}\) is y – y₁ = m(x – x₁)
or, y – 2 = \(\frac{4}{3}\) (x + 4)
or, 3y – 6 = 4x + 16
or, 4x – 3y + 22 = 0

(k) Passing through the point (a cos³ θ, a sin³ θ) and perpendicular to the straight line x sec θ + y cosec θ = α.
Solution:
The slope of the line x sec θ + y cosec θ = a is \(\frac{-\sec \theta}{{cosec} \theta}\) = -tanθ
∴ Slope of the required line  = cot θ
∴ Equation of the line through (a cos³ θ, a sin³ θ) is y – y₁ = m(x – x₁)
or, y – a sin³ θ = cot θ(x – a cos³ θ)
or, y – a sin³ θ = \(\frac{\cos \theta}{\sin \theta}\) (x – a cos³ θ)
or y sin θ – a sin⁴ θ = x cos  θ – a cos⁴ θ
or (x cos θ – y sin θ) + a(sin⁴ θ – cos⁴ θ) = 0

(l) Which passes through the point (3, -4) and is such that its portion between the axes is divided at this point internally in the ratio 2: 3.
Solution:

(m) which passes through the point (α, β) and is such that the given point bisects its portion between the coordinate axis.
Solution:

x = 2α , y = 2β
∴ Equation of the line \(\overleftrightarrow{\mathrm{AB}}\) is \(\frac{x}{2 \alpha}+\frac{y}{2 \beta}\) = 1(Intercept form)

Question 7.
(a) Find the equation of the lines that is parallel to the line 3x + 4y + 7 = 0 and is at a distance 2 from it.
Solution:
3x + 4y + 7 = 0
or, \(\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}\) = 0(Normal form)
∴ Equation of the lines parallel to the above line and 2 units away from it are \(\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}\) ± 2 = 0
or, 3x + 4y + 7 ± 10 = 0
∴ 3x + 4y + 17 = 0 and 3x + 4y – 3 = 0

(b) Find the equations of diagonals of the parallelogram formed by the lines ax + by = 0, ax + by + c = 0, lx + my = 0, and lx + my + n = 0. What is the condition that this will be a rhombus?
Solution:

(c) Find the equation of the line passing through the intersection of 2x – y – 1 = 0 and 3x – 4y + 6 = 0 and parallel to the line x + y – 2 = 0.
Solution:
Let the equation of the required line be (2x – y – 1) + λ(3x – 4y + 6) = 0
or, x(2 + 3λ) + λ(-1 – Aλ) + 6λ – 1 = 0
As this line is parallel to the line x + y – 2 = 0
we have their slopes are equal.
∴ \(-\left(\frac{2+3 \lambda}{-1-4 \lambda}\right)=\frac{-1}{1}\)
or, 2 + 3λ = -1 – 4λ
or, 7λ = -3 or, λ = \(\frac{-3}{7}\)
∴ Equation of the line is (2x – y – 1) – \(\frac{3}{7}\) (3x – 4y + 6) = 0
or, 14x – 7y – 1 – 9x + 12y – 18 = 0
or, 5x + 5y – 25= 0
or, x + y = 5

(d) Find the equation of the line passing through the point of intersection of lines x + 3y + 2 = 0 and x – 2y – 4 = 0 and perpendicular to the line 2y + 5x – 9 = 0.
Solution:
Let the equation of the line be (x + 3y + 2) + λ(x – 2y – 4) = 0
or, x(1 + λ) + y(3 – 2λ) + 2 – 4λ = 0
As this line is perpendicular to the line 2y + 5x – 9 = 0.
We have the product of their slopes is -1.
∴ \(\frac{1+\lambda}{3-2 \lambda} \times \frac{5}{2}\) = -1
or, 5 + 5λ = – 6 + 4λ
or, λ = -6 – 5 = -11.
∴ Equation of the required line is (x + 3y + 2) – 11(x – 2y – 4) = 0
or, x + 3y + 2- 11x + 22y + 44 = 0
or, – 10x + 25y + 46 = 0
or, 10x – 25y – 46 = 0

(e) Find the equation of the line passing through the intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0 and the centroid of the triangle whose vertices are the points (3, -1) (1, 3) and (2, 4).
Solution:
Let the equation of the required line (x + 3y – 1) + λ(3x – y + 1) = 0   … (1)
Again, the centroid of the triangle with vertices (3, – 1), (1, 3), and (2, 4) is \(\left(\frac{3+1+2}{3}, \frac{-1+3+4}{3}\right)\) = (2, 2)
As line (1) passes through (2, 2), we have (2 + 6 – 1) +1(6 – 2 + 1) = 0
or, 7 + 5λ = 0 or, λ = \(\frac{-7}{5}\)
∴ Equation of the line (x + 3y – 1) – \(\frac{7}{5}\) (3x – y + 1) = 0
or, 5x + 15y – 5 – 21x + 7y – 7 = 0
or, 22y – 16x – 12 = 0
or, 11y – 8x – 6 = 0
or, 8x – 11y + 6 = 0

Question 8.
If lx + my + 3 = 0 and 3x – 2y – 1 = 0 represent the same line, find the values of l and m.
Solution:
lx + my + 3 = 0 and 3x – 2y – 1 = 0 represents the same line
∴ \(\frac{l}{3}=\frac{m}{-2}=\frac{3}{-1}\)
∴ l = -9, m = 6

Question 9.
Find the equation of sides of a triangle whose vertices are at (1, 2), (2, 3), and (-3, -5).
Solution:
Equation of \(\overline{\mathrm{AB}}\) is \(y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\)
\(y-2=\frac{3-2}{2-1}(x-1)\)
or, y – 2 = x – 1
or, x – y + 1 = 0

Question 10.
Show that origin is within the triangle whose sides are given by equations, 3x – 2y = 1, 5x + 3y + 11 = 0, and x – 7y + 25 = 0.
Solution:



∴ The origin lies within the triangle ABC.

Question 11.
(a) Find the equations of straight lines passing through the point (3, -2) and making an angle 45° with the line 6x + 5y = 1.

or, 11x – y = 35, x + 11y + 19 = 0

(b) Two straight lines are drawn through the point (3, 4) inclined at an angle 45° to the line x – y – 2 = 0. Find their equations and obtain area included by the above three lines.
Solution:

Slope of L₂ = 0
Then as L₂ ⊥ L₃
Slope of L₃ = ∞
∴ Equation of L₂ is
y – y₁ =m(x – x₁)
or, y – 4 = 0(x – 3) = 0
or, y = 4
∴ Equation of L₃ is y – 4 = ∞ (x – 3)
or, x – 3 = 0 or, x = 3
∴ Sloving L₁ and L₂, we have
x – y – 2 = 0, y = 4
or, x = 6
The coordinates of A are (6, 4).
Again solving L1 and L3, we have
x – y – 2 = 0, x = 3
or, y = x – 2 = 3 – 2 = 1
∴ The coordinates of B are (3, 1).
Area of the triangle PAB is

(c) Show that the area of the triangle formed by the lines given by the equations y = m₁x + c₁,y = m₂x + c_2, and x = 0 is \(\frac{1}{2} \frac{\left(c_1-c_2\right)^2}{\left[m_2-m_1\right]}\)
Solution:

Question 12.
Find the equation of lines passing through the origin and perpendicular to the lines 3x + 2y – 5 = 0 and 4x + 3y = 7. Obtain the coordinates of the points where these perpendiculars meet the given lines. Prove that the equation of a line passing through these two points is 23x + 11y – 35 = 0.
Solution:
The slopes of the line 3x + 2y – 5 = 0 and 4x + 3y = 7 are \(\frac{-3}{2}\) and \(\frac{-4}{3}\)
∴ Slopes of the lines perpendicular to the above lines are \(\frac{2}{3}\) and \(\frac{3}{4}\)
∴ Equation of the lines through the origin and having slopes \(\frac{2}{3}\) and \(\frac{3}{4}\)
y = \(\frac{2x}{3}\) and y = \(\frac{3x}{4}\)
Now solving 3x + 2y – 5 = 0 and y = \(\frac{2x}{3}\)
we have 3x + \(\frac{4x}{3}\) – 5 = 0
or, 9x + 4x – 15 = 0
or, x = \(\frac{15}{13}\)
∴ y = \(\frac{2 x}{3}=\frac{2}{3} \times \frac{15}{13}=\frac{10}{13}\)
∴ The perpendicular y = \(\frac{2 x}{3}\) meets the line 3x + 2y – 5 = 0 at \(\left(\frac{15}{13}, \frac{10}{13}\right)\)
Again, solving 4x + 3y = 7 and y = \(\frac{3 x}{4}\)
we have 4x + 3 × \(\frac{3 x}{4}\) = 7
or, 16x + 9x = 28 or, x = \(\frac{28}{25}\)

Question 13.
(a) Find the length of a perpendicular drawn from the point (-3, -4) to the straight line whose equation is 12x – 5y + 65 = 0.
Solution:
The length of the perpendicular drawn from the point (- 3, -4) to the straight line 12x – 5y + 65 = 0 is

(b) Find the perpendicular distances of the point (2, 1) from the parallel lines 3x – 4y + 4 = 0 and 4y – 3x + 5 = 0. Hence find the distance between them.
Solution:
The distance of the point (2, 1) from the line 3x – 4y + 4 = 0 is \(\left|\frac{3 \times 2-4 \times 1+4}{\sqrt{9+16}}\right|=\frac{6}{5}\)
Again distance of the point (2, 1) from the line 4y – 3x + 5 = 0 is

(c) Find the distance of the point (3, 2) from, the line x + 3y – 1 = 0, measured parallel to the line 3x – 4y + 1 = 0
Solution:

Let the coordinates of M be (h, k).
As \(\overline{\mathrm{PM}} \| \mathrm{L}_1\), We have their slopes are equal.

(d) Find the distance of the point (-1, -2) from the line x + 3y – 7 = 0, measured parallel to the line 3x + 2y – 5 = 0
Solution:
Slope of the line 3x + 2y – 5 = 0 is \(\left(-\frac{3}{2}\right)\)
Equation of the line through (-1, -2) and parallel to this line is y + 2 = – \(\frac{3}{2}\) (x + 1)
⇒ 2y + 4 = -3x – 3
⇒ 3x + 2y + 7 = 0 …(1)
Given line is : x + 3y – 7 = 0    ….(2)
from (1) and (2) we get 7y – 28 = 0 Py = 4 and x = \(-\frac{35}{7}\) = -5
Thus the required distance is \(\sqrt{(-1+5)^2+(-2-4)^2} \quad=\sqrt{16+36}\)
= √52 = 2√3 units.

(e) Fine the distance of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) from the origin.
Solution:
The equation of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) x – x₁
or, y – a sin α

Question 14.
Find the length of perpendiculars drawn from the origin on the sides of the triangle whose vertices are A( 2, 1), B (3, 2), and C (- 1, -1).
Solution:

Question 15.
Show that the product of perpendicular from the points \(\left(\pm \sqrt{a^2-b^2}, 0\right)\) upon the straight line \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 is b²
Solution:

Question 16.
Show that the lengths of perpendiculars drawn from any point of the straight line 2x + 11y – 5 = 0 on the lines 24x + 7y – 20 = 0 and 4x – 3y – 2 = 0 are equal to each other.
Solution:
Let P(h, k) is any point on the line
2k + 11y – 5 = 0
∴ 2h + 11k – 5 = 0
Now the length of the perpendicular from P on the line 24x + 7y – 20 = 0 is

Clearly d₁ = d₂

Question 17.
If p and p’ are the length of perpendiculars drawn from the origin upon the lines x sec α + y cosec α = 0 and x cos α – y sin α – a cos 2α = 0
Prove that 4p² + p’² = a²
Solution:

Question 18.
Obtain the equation of the lines passing through the foot of the perpendicular from (h, k) on the line Ax + By + C = 0 and bisect the angle between the perpendicular and the given line.
Solution:

Slope of the line L is \(\frac{-\mathrm{A}}{\mathrm{B}}\)
∴ Slope of the line \(\overline{\mathrm{PM}} \text { is } \frac{\mathrm{B}}{\mathrm{A}}\)
∴ Equation of the line \(\overline{\mathrm{PM}}\) is y – y₁ = m(x – x₁)
or, y – k = \(\frac{\mathrm{B}}{\mathrm{A}}\) (x – h)
or, Ay – Ak =Bx – Bh
or, Bx – Ay + Ak – Bh = 0
∴ Equation of the bisectors of the angles between the lines L and \(\overline{\mathrm{PM}}\) is
\(\frac{\mathrm{A} x+\mathrm{B} y+\mathrm{C}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\pm \frac{\mathrm{B} x-\mathrm{A} y+\mathrm{A} k-\mathrm{B} h}{\sqrt{\mathrm{B}^2+\mathrm{A}^2}}\)
or, Ax + By + C = ± (Bx – Ay +Ak – Bh)

Question 19.
Find the direction in which a straight line must be drawn through the point(1, 2) such that its point of intersection with the line x + y – 4 = 0 is at a distant \(\frac{1}{3} \sqrt{6}\) from this point.
Solution:

Question 20.
A triangle has its three sides formed by the lines x + y = 3, x + 3y = 3, and 3x + 2y = 6. Without solving for the vertices, find the equation of its altitudes and also calculate the angles of the triangle.
Solution:

or 1 + 3λ = – 3 – 6λ
or 9λ = – 4 or , λ = \(\frac{-4}{9}\)
∴ Equation of \(\overline{\mathrm{AD}}\) is (x + y – 3) – \(\frac{4}{9}\) (3x + 2y – 6) = 0
or, 9x + 9y – 27 – 12x – 8y + 24 = 0
or, -3x + y – 3 = 0
or, 3x – y + 3 = 0
Let the equation of \(\overline{\mathrm{BE}}\) be (x + y – 3) + 1(x + 3y – 3) = 0
or, x(1 + λ) + y( 1 + 3λ) – 3 – 3λ = 0
As \(\overline{\mathrm{BE}} \perp \overline{\mathrm{AC}}\)
we have \(\frac{1+\lambda}{1+3 \lambda} \times \frac{3}{2}\) = -1
or, 3 + 3λ = -2 – 6λ
or, 9λ = – 5 or λ = \(\frac{-5}{9}\)
∴ Equation of \(\overline{\mathrm{BE}}\) is (x + y – 3) – \(\frac{5}{9}\) (x + 3y – 3) = 0
or 9x + 9y – 27 – 5x – 15y + 15 = 0
or, 4x – 6y – 12 = 0
or, 2x – 3y – 6 = 0
Let the equation of \(\overline{\mathrm{CE}}\) be (3x + 2y – 6) + λ (x + 3y – 3) = 0
x(3 + λ) + y (x + 3λ) – 6 – 3λ = 0
As \(\overline{\mathrm{CF}} \perp \overline{\mathrm{AB}} .\)
we have \(\frac{3+\lambda}{2+3 \lambda}\) × 1 = -1
or, 3 + λ = -2 – 3λ
or, 4λ = -2 – 3 = -5
or, λ = \(\frac{-5}{4}\)
∴ Equation of is \(\overline{\mathrm{CF}}\) (3x + 2y – 6) \(\frac{-5}{4}\) (x + 3y – 3) = 0
or, 12x + 8y – 24 – 5x – 15y + 15 =0
or, 7x – 7y – 9 = 0

Question 21.
A triangle has its vertices at P(1, -1), Q(3, 4) and R(2, 5). Find the equation of altitudes through P and Q and obtain the coordinates of their point of intersection. (This point is called the ortho-center of the triangle.)
Solution:

Question 22.
(a) Show that the line passing through (6, 0) and (-2, -4) is concurrent with the lines
2x – 3y – 11 = 0 and 3x – 4y = 16
Solution:
The equation of the line through (6,0) and (-2, -4) is

(b) Show that the lines lx + my + n = 0 mx + ny + 1 = 0 and nx + ly + m = 0 are concurrent, l + m + n = 0
Solution:
As the lines
lx + my + n = 0
mx + ny + 1 = 0
and nx + ly + m = 0 are concurrent.

Question 23.
Obtain the equation of the bisector of the acute angle between the pair of lines.
(a) x + 2y = 1, 2x + y + 3 = 0
Solution:
Equation ofthe bisectors ofthe angles between the lines x + 2y – 1 = 0 and 2x + y + 3 = 0 are \(\frac{x+2 y-1}{\sqrt{1^2+2^2}}=\pm \frac{2 x+y+3}{\sqrt{2^2+1^2}}\)
or, x + 2y – 1 = ± (2x + y + 3)
∴ x + 2y – 1 = 2x + y + 3 and
x + 2y – 1= -2x – y – 3
∴ x – y + 4 = 0 and 3x + 3y + 2 = 0
Let θ be the angle between x + 2y – 1 = 0 and x – y + 4 = 0
∴ tan θ = \(\frac{a_1 b_2-a_2 b_1}{a_1 a_2+b_1 b_2}\)
\(=\frac{1 \cdot(-1)-(+1) \cdot 2}{1 \cdot 1+2(-1)}\)
\(=\frac{-1-2}{1-2}=\frac{-3}{-1}=3\)
sec² θ = 1 + tan² θ = 1 + 9 = 10
cos² θ = 1/10
cos θ = \(\frac{1}{\sqrt{10}}<\frac{1}{\sqrt{2}}\) ⇒ θ > 45°
∴ x – y + 4 = 0 is the obtuse angle bisector.
⇒ 3x + 3y + 2 = 0 is acute angle bisector.

(b) 3x – 4y = 5, 12y – 5x = 2
Solution:
Given equation of lines are
3x – 4y – 5 = 0    …..(1) and  5x – 12y + 2 = 0     ……(2)
Equation of bisectors of angles between these Unes are:

Question 24.
(a) Find the coordinates of the center of the inscribed circle of the triangle formed by the line x cos α + y sin α = p with the coordinate axes.
Solution:


\(\left(\frac{p}{\sin \alpha+\cos \alpha+1}, \frac{p}{\sin \alpha+\cos \alpha+1}\right)\)

(b) Find the coordinates of the circumcentre and incentre of the triangle formed by the lines 3x – y = 5, x + 2y = 4, and 5x + 3y + 1 = 0.
Solution:

Question 25.
The vertices B, and C of a triangle ABC lie on the lines 3y = 4x and y = 0 respectively, and the side \(\overline{\mathbf{B C}}\) passes through the point (2/3, 2/3). If ABOC is a rhombus, where O is the origin, find the equation of \(\overline{\mathbf{B C}}\) and also the coordinates of A.
Answer:
Let the coordinates of C be (a, 0) so that the length of the side of the rhombus is ‘a’

Question 26.
Find the equation of the lines represented by the following equations.
(a) 4x² – y² = 0
Solution:
4x² – y² = 0
or, (2x + y)(2x – y) = 0
∴ 2x + y = 0 and 2x – y = 0 are the two separate lines.

(b) 2x² – 5xy – 3y²
Solution:
2x² – 5xy – 3y²
or, 2x² – 6xy + xy – 3y² = 0
or, 2x(x – 3y) + y(x – 3y) = 0
or, (x – 3y)(2x + y) = 0
∴ x – 3y = 0 and 2x + y = 0 are the two separate lines.

(c) x² + 2xy sec θ + y² = 0
Solution:
x² + 2xy sec θ + y² = 0
∴ a = 1, b = 2y sec θ, c = y²
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{2 y \sec \theta \pm \sqrt{4 y^2 \sec ^2 \theta-4 y^2}}{2}\)
= \(\frac{-2 y \sec \theta \pm 2 y \tan \theta}{2}\)
= y (- sec θ ± tan θ)
∴ x = y (- sec θ + tan θ) and x – y (- sec θ – tan θ) are the two separate lines.

(d) 3x² + 4xy = 0
Solution:
3x² + 4xy = 0
or x (3x + 4y) = 0
∴ x = 0 and 3x + 4y = 0 are the two separate lines.

Question 27.
From the equations which represent the following pair of lines.
(a) y = mx; y = nx
Solution:
y – mx = 0, y – nx = 0
or (y – mx) (y – nx) = 0
or, y² – nxy – mxy + mnx² = 0
or, y² – xy (m + n) + mnc² = 0 which is the equation of a pair of lines.

(b) y – 3x = 0 ; y + 3x = 0
Solution:
y – 3x = 0, y + 3x = 0
∴ (y – 3x) (y + 3x) = 0
or, y² – 9x² = 0 which is the equation of a pair of lines.

(c) 2x – 3y + 1 = 0 ; 2x + 3y + 1 = 0
Solution:
2x – 3y + 1 = 0; 2x + 3y + 1 =0
or, (2x – 3y + 1)(2x + 3y + 1) = 0
or, (2x + 1)² – 9y² = 0
or, 4x² + 1 + 4x – 9y² = 0
or, 4x² – 9y² + 4x + 1= 0 which represents a pair of lines.

(d) x = y. x + 2y + 5 = 0
Solution:
x = y, x + 2y + 5 = 0
∴ (x – y) (x + 2y + 5) = 0
or, x² + 2xy + 5x – xy – 2y² – 5y =0
or, x² – 2y² + xy + 5x – 5y = 0 which represents a pair of lines.

Question 28.
Which of the following equations represents a pair of lines?
(a) 2x² – 6y² + 3x +  y + 1 = 0
Solution:
a = 2, b = -6, 2g = 3
2f = 1, c = 1
∴ g = \(\frac{3}{2}\), f = \(\frac{1}{2}\), h = 0
∴ abc + 2fgh – ah² – bg² – ch²
= 2(-6). 1 + 2 × \(\frac{1}{2}\) × \(\frac{3}{2}\) – 0 – (-6) × \(\frac{9}{4}\) – 1 × 0
= -12 + \(\frac{3}{2}\) + \(\frac{27}{2}\) = \(\frac{6}{2}\) = 3 ≠ 0
∴ The given equation does not represent a pair or lines.

(b) 10x² – xy – 6y² – x + 5y – 1 = 0
Solution:
a = 10. 2h = 1
B = -6, 2g = -1
2f = 5. C= -1
∴ h = –\(\frac{1}{2}\), g = –\(\frac{1}{2}\) , f = \(\frac{5}{2}\)
∴ abc + 2fgh – ah² – bg² – ch²
= 10(-6)(-1) + 2 × \(\frac{5}{2}\) × (-\(\frac{1}{2}\)) × (-\(\frac{5}{2}\)) – 10 × \(\frac{2.5}{4}\) – (-6)\(\frac{1}{4}\) – (-1)\(\frac{1}{4}\)
= 60 + \(\frac{5}{4}\) – \(\frac{250}{4}\) + \(\frac{6}{4}\) + \(\frac{1}{4}\)
= \(\frac{240+5+6-250+1}{4}=\frac{2}{4}\)
∴ The given equation does not represent a pair of lines.

(c) xy + x + y + 1 = 0
Solution:
xy + x + y + 1= 0
or, x(y + 1) + 1(y + 1 ) = 0
or (y + 1 )(x + 1) =0
∴ x + 1 = 0
and y + 1 = 0 are the two separate lines,
∴ The given equation represents a pair of lines.

Question 29.
For what value of λ do the following equations represent pair of straight lines?
(a) λx² + 5xy – 2y² – 8x + 5y – λ = 0
Solution:
λx² + 5xy – 2y² – 8x + 5y – λ = 0
∴ a = λ, 2h = 5, b = -2, 2g = -8
2f = 5, c = -1
∴ h = \(\frac{5}{2}\), g = -4, f = \(\frac{5}{2}\)
As the given equation represent a pair of lines, we have abc + 2fgh – ah² – bg² – ch² = 0
or, λ(-2)(-λ) + 2. \(\frac{5}{2}\) (-4). \(\frac{5}{2}\) -λ × \(\frac{25}{4}\) – (-2) (-4)² – (-λ) × \(\frac{25}{4}\) = 0
or, 2λ² – 50 – \(\frac{25 λ }{4}\) + 32 + \(\frac{25 λ }{4}\) = 0
or, 2λ² = 18 or, λ² = 9
λ = ±3

(b) x² – 4xy – y² +6x + 8y + λ = 0
Solution:
Here a = 1, 2h = -1, b = -1, 2g = 6, 2f = 8, c = τ
As the given equation represent a pair of lines, we have
abc + 2fgh – af² – bg² – ch² = 0
⇒ (-1) τ + 2.4.3 (-2) – 1. 4² – (-1). ³2 – τ(-2)² = 0
⇒ -τ – 48 – 16 + 9 – 4τ = 0
⇒ -5τ – 55 = 0 ⇒ τ = -11

Question 30.
(a) Obtain the value of λ for which the pair of straight lines represented by 3x² – 8xy + λy² = 0 are perpendicular to each other.
Solution:
3x² – 8xy + λy² = 0
∴ a = 3. 2h = -8, b = λ
As the pair of lines are perpendicular to each other, we have a + b = 0.
or, 3 + λ = 0 – or, λ = -3

(b) Prove that a pair of lines through the origin perpendicular to the pair of lines represented by px² – 2qxy + ry² = 0 is given by rx² – 2qxy + py² = 0
Solution:
px² – 2qxy + ry² = 0
∴ a = p, b = 2qy, c = ry²

(c) Obtain the condition that a line of the pair of lines ax² + 2hxy + by² = 0,
(i) Coincides with
Solution:

(ii) is perpendicular to, a line of the pair of lines px² + 2qxy + ry² = 0
Solution:

Question 31.
Find the acute angle between the pair of lines given by :
(a) x² + 2xy – 4y² = 0
Solution:
x² + 2xy – 4y² = 0
∴ a=1, 2h = 2, b = -4
∴ tan θ = \(\frac{\pm 2 \sqrt{h^2-a b}}{a+b}=\frac{\pm 2 \sqrt{1+4}}{1-4}\)
\(=\pm \frac{2 \sqrt{5}}{-3}=\mp \frac{2 \sqrt{5}}{3}\)
∴ The acute angle between the pair of lines is tan^-1 \(\frac{2 \sqrt{5}}{3}\)

(b) 2x² + xy – 3y² + 3x + 2y + 1 = 0
Solution:
2x² + xy – 3y² + 3x + 2y + 1 = 0
∴ a = 2, 2h = 1, b = -3, 2g = 3 2f= 2, c = 1.
tan θ = \(\frac{\pm 2 \sqrt{h^2-a b}}{a+b}\)
\(=\pm \frac{2 \sqrt{\frac{1}{4}+6}}{2-3}=\pm \frac{2 \times 5}{2(-1)}\) = ± 5
∴ The acute angle is tan^-1 5

(c) x² + xy – 6y² – x – 8y – 2 = 0
Solution:
Given Equation is x² + xy – 6y² – x – 8y – 2 = 0
here a = 1, 2h = 1, b = -6 thus if 0 is the acute angle between two lines then
tan θ = \(=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|\)
= \(\left|\frac{2 \times 5}{-10}\right|\) = 1
∴ θ = 45°

Question: 32.
Write down the equation of the pair of bisectors of the following pair of lines :
(a) x² – y² = 0 ;
Solution:
x² – y² = 0
∴ a = 1, b = -1, h = 0
∴ The equation of the bisectors of the angles between the pair of lines are \(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{1+1}=\frac{x y}{0}\)
or, xy = 0

(b) 4x² – xy – 3y² = 0
Solution:
4x² – xy – 3y² = 0
∴ a = 4, 2h = -1, b = -3
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{(a-b)}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{7}=\frac{x y}{\left(-\frac{1}{2}\right)}\)
or, x² – y² = -14xy
or, x² + 14xy – y² = 0

(c) x² cos θ + 2xy – y² sin θ = 0
Solution:
x² cos θ + 2xy – y² sin θ = 0
∴ a = cos θ, 2h = 2, b = – sin θ
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{\cos \theta+\sin \theta}=\frac{x y}{1}\)
or, x² – y² = xy(cos θ + sin θ)

(d) x² – 2xy tan θ – y² = 0
Solution:
x² – 2xy tan θ – y² = 0
∴ a = 1, 2h = -2 tan θ, b = -1
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{2}=\frac{x y}{-\tan \theta}\)
or, x² – y² = 2xy cot θ
or, x² + 2xy cot θ – y² = 0

Question 33.
If the pair of lines represented by x² – 2pxy – y² = 0 and x² – 2qxy – y² = 0 be such that each pair bisects the angle between the other pair, then prove that pq = -1.
Solution:

Question 34.
Transform the equation: x² + y² – 2x – 4y + 1 = 0 by shifting the origin to (1, 2) and keeping the axes parallel.
Solution:
x² + y² – 2x – 4y + 1 = 0     ……(1)
Let h = 1, k = 2
Taking x’ + h and y = y’ + k we have
(x’ + h)² + (y’ + k)² – 2(x’ + j) -4(y’ + k) + 1=0
or, (x + 1)² + (y’ + 2)² – 2(x’ + 1) – 4(y’+ 2) + 1=0
or, x^‘2 + 1 + 2x’ + y^‘2 + 4 – 4y’ – 2x’ – 2 – 4y’- 8 + 1 = 0
or, x^‘2 + y’² – 4 = 0
∴ The transformed equation is x² + y² = 4

Question 35.
Transform the equation: 2x² + 3y² + 4xy – 12x – 14y + 20 = 0. When referred to parallel axes through(2, 1).
Solution:
2x² + 3y² + 4xy – 12x – 14y + 20 = 0
Let h = 2, k = 1
Taking x = x’ + 1 and y = y’ + 1
we have
2(x’ + 2)² + 3(y’ + 1)² + 4(x’ + k)(y’ + 1) – 12 (x’ + 2)- 14 (y’ + 1) + 20 = 0
or, 2x^‘2 + 8 + 8x’ + 3 + 6y’ + 3y^’2 + 4x’y’ + 4x’ + 8y’ + 8 – 12x’ – 14y’ – 18 = 0
or, 2x^‘2 + 3y^’2 + 4x’y’ + 1=0
The transformed equation is
2x² + 3y² + 4xy + 1 = 0

Question 36.
Find the measure of rotation so that the equation x² – xy + y² = 5 when transformed does not contain xy- term.
Solution:
x² – xy + y² = 5
Taking x = x’ cos α – y’ sin α
y = x’ sin α – y’ cos α
We get (x’ cos α – y’ sin α)² – (x’ cos α – y’ cos α) (x’ sin α + y’ cos α) + (x’ sin α + y’ cos α)² = 5
⇒ x^‘2 cos² α + y^‘2 sin² α – 2x’y sin α.
cos α – x^‘2 sin α. cos α – x’y cos² α + x’y’ sin² α + y^‘2 sin α. cos α + x^‘2 sin² α + y^‘2 cos² α + 2x’y’ sin α cos α = 5
Given that the transformed equation does not xy term.
Hence the co-efficient of x’y’ is zero.
That is sin² α – cos² α = 0
⇒ sin² α = cos² α
⇒ tan² α = 1 ⇒ tan α = 1 ⇒ α= 45°

Question 37.
What does the equation x + 2y – 10 =0 become when the origin is changed to (4, 3)?
Solution:
x + 2y – 10 = 0
Let h = 4, k = 3
Taking x = x’ + 4, y = y’ + 3
we have x’ + 4 + 2 (y’ + 3) – 10 = 0
or, x + 2y’ = 0
∴ The transformed equation is x + 2y = 0.

Odisha State Board BSE Odisha 10th Class Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(a) Textbook Exercise Questions and Answers.

BSE Odisha Class 10 Maths Solutions Geometry Chapter 1 ଜ୍ୟାମିତିରେ ସାଦୃଶ୍ୟ Ex 1(a)

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(a) ବିମ୍ନ ଟିତ୍ର (a) 6ର L₁ || L₂ || L₃ ଏର୍ଚ T₁ ଓ T₂ ଛେଦ ଦା |
(i) AB = 2 6ସ.ର୍ମି., BC = 3 6ସ.ର୍ମି. ଓ DE = 3 6ସ.ର୍ମି. 6ହ6ଲ EF = ……….|
(ii) DE = 6 6ସ.ର୍ମି., EF = 8 6ସ.ର୍ମି. ଓ BC = 6 6ସ.ର୍ମି. 6ହ6ଲ AC = ……….|

(b) ଗପଭୋକ୍ତ ତିତ୍ର (b) 6ର L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଛେଦକ |
(i) AB = 1.5 × BC ହେଲେ, \(\frac { EF }{ FD }\) = ……………..
(ii) \(\overline{\mathrm{AC}}\) ର ମଧ୍ୟବିରୁ B 6ହ6ଲ, EF ର ……… ମୁଶ 6ହରଛି FD|
Solution:
(a) (i) L₁ || L₂ || L₃ ⇒ \(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\)
⇒ \(\frac { 2 }{ 3 }\) = \(\frac { 3 }{ EF }\) ⇒ EF = \(\frac{3 \times 3}{2}\) 6 ସ.ମି. = 4.5 6ସ.ମି.

(ii) L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ରୁକଚି 6ଛଦକ |
∴ \(\frac { DE }{ EF }\) = \(\frac { AB }{ BC }\) ⇒ \(\frac { 6 }{ 8 }\) = \(\frac { AB }{ 6 }\)
⇒ AB = \(\frac{6 \times 6}{8}\) = 4.5 6ସ.ମି. |
AC = AB + BC = 4.5 6ସ.ମି. + 6ସ.ମି. = 10.5 6ସ.ମି. |

(b) (i) ଏଠ।6ର L₁ || L₂ || L₃ ଏରଂ T₁ ଓ T₂ ଦୁଲଟି 6ଚ୍ଚଦକ |
\(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac { 3 }{ 2 }\) = \(\frac { DE }{ EF }\)
⇒ \(\frac{3+2}{2}\) = \(\frac{\mathrm{DE}+\mathrm{EF}}{\mathrm{EF}}\) ⇒ \(\frac { DE }{ EF }\) = \(\frac { 5 }{ 2 }\) ⇒ \(\frac { EF }{ DF }\) = \(\frac { 2 }{ 5 }\)

(ii) \(\overline{\mathrm{AC}}\) ର ମଧ୍ୟ ଦିନୁ B ଅର୍ଥ।ତ୍ AB = BC
\(\frac { EF }{ FD }\) = \(\frac { BC }{ AC }\) = \(\frac{\mathrm{BC}}{\mathrm{AB}+\mathrm{BC}}\) = \(\frac { BC }{ 2BC }\) = \(\frac { 1 }{ 2 }\) ⇒ 2EF = FD>
ଅର୍ଥାତ୍ EF ଭ 2 ଣ୍ଣଣ 6ଦ୍ରଣଛି FD |

Question 2.
ଟିତ୍ର6ର L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଦୁଲଟି 6ଚ୍ଛଦକ | L₂ ଓ L₃ ରପ6ର ପ୍ଥଥ।କୁ6ର G ଓ H ଦିହୁ ରିଜିତ 6 ପ୍ପପତି BC = AD ଏବଂ CH = BE;

ପ୍ରମାଣ କର 6ଯ
(i) DG : EH = DE : EF
(ii) (DG + EH) : EH = DF : EF
Solution:
ଦଉ : L₁ || L₂ || L₃ ଏବଂ T₁ ଓ T₂ ଦୁକଟି 6ଚ୍ଛଦକ ଯଥାସ୍ତ6ମ L₁, L₂ ଓ L₃ କୁ A,B,C ଓ D,E,F ଦିନ୍ଦ6ର 6ଚ୍ଛଦକ6ର | L₂ ରପ6ର G ଏକ ଦିବ୍ର 6ସ୍ ପରି BG = AD ଏର୍ବ L₃ ରପ6ର H ଏକ ଦିନ୍ଦୁ 6ନ୍ଦ୍ ପରି CH = BE |
ପ୍ରାମାଣ୍ୟ :
(i) DG : EH = DE : EF
(ii) (DG + EH) : EH = DF : EF

ଅକ୍ଳଜ : \(\overline{\mathrm{DG}}\) ଓ \(\overline{\mathrm{EH}}\) ଅଲ୍ଲବ ଦବାପାର |
ପ୍ତମାଣ : (i) AD = BG (ଦର) \((\overline{\mathrm{AD}} \| \overline{\mathrm{BG}})\), ∵ L₁ || L₂ (ଦର)
⇒ ABGD ଏକ ସାମାତ୍ରତିଦ ଚିତ୍ର | AB = DG
6ସଦୃପତି BEHC ଏକ ସାମାତ୍ର ଚିତ୍ର ଓ BC = EH
∴ \(\frac { DG }{ EH }\) = \(\frac { AB }{ BC }\) …(1)
ପୁନଣ୍ଡ \(\frac { AB }{ BC }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac { DG }{ EH }\) = \(\frac { DE }{ EF }\) (1ରୁ) (ପ୍ରମାଣିତ)

(ii) ପୁଫରୁ ପ୍ରମାଣିର \(\frac { DG }{ EH }\) = \(\frac { DE }{ EF }\) ⇒ \(\frac{\mathrm{DG}+\mathrm{EH}}{\mathrm{EH}}\) = \(\frac{\mathrm{DE}+\mathrm{EF}}{\mathrm{EF}}\) (Componendo ଦ୍ଦାରା )
⇒ \(\frac{\mathrm{DG}+\mathrm{EH}}{\mathrm{EH}}\) = \(\frac { DF }{ EF }\) (ପ୍ରମାଣିତ)

Question 3.
ନିମ୍ନ ଚିତ୍ରରେ L₁ || L₂ || L₃ ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ । ଯଦି AB = BC ହୁଏ, ପ୍ରମାଣ କର ଯେ 2 BE = AD + CF |
Solution:
ଦତ୍ତ : L₁ || L₂ || L₃ ଛେଦକ T₁ ଓ T₂ , L₁, L₂, ଓ L₃, କୁ ଯଥାକ୍ରମେ A, B, C ଓ D, E, F ବିନ୍ଦୁରେ ଛେଦ କରେ ଏବଂ AB = BC |
ପ୍ରାମାଣ୍ୟ : 2BE = AD + CF
ଅକନ : E ଦିଦୁ ମଧ୍ୟ6ଦକ AC ସମାତ ତା6ଦ ଅଜିତ ବେଡା L₁ || L₃ କୁ ଯଥାକୁ6ମ X ଓ Y ଦିଦ6ର 6ଛଦକ୍ଳରୁ |

ଅଙ୍କନ : E ବିନ୍ଦୁ ମଧ୍ୟଦେଇ \(\overline{\mathrm{AC}}\)ସହ ସମାନ୍ତର ଭାବେ ଅଙ୍କିତ ରେଖା L₁ ଓ L₃, କୁ ଯଥାକ୍ରମେ X ଓ Y ବିନ୍ଦୁରେ ଛେଦକରୁ ।

Question 4.
ଚିତ୍ରରେ L₁ || L₂ || L₃, ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ । L₁, L₂, ଓ L₃, କୁ ଛେଦକ T₁, ଯଥାକ୍ରମେ A, B ଓ C ବିନ୍ଦୁରେ ଛେଦକରେ ଏବଂ L₁, L₂, ଓ L₃, କୁ ଛେଦକ T₂, ଯଥାକ୍ରମେ D, E ଓ F ବିନ୍ଦୁରେ ଛେଦ କରେ । DE = EF ହେଲେ, ପ୍ରମାଣ କର ଯେ, CF – AD = 2 EB । (ସୂଚନା : AF ଅଙ୍କନ କର)

Solution:
ଦତ୍ତ : L₁ || L₂ || L₃ ଏବଂ T₁, ଓ T₂, ଦୁଇଟି ଛେଦକ ଯଥାକ୍ରମେ L₁, L₂, ଓ L₃, କୁ A, B, C ଓ D, E, F ବିନ୍ଦୁରେ ଛେଦକରେ ।
ପ୍ରାମାଣ୍ୟ : CF – AD = 2EB.
ଅଙ୍କନ : \(\overline{\mathrm{AF}}\) ଅଙ୍କନ କର । B ବିନ୍ଦୁରୁ \(\overline{\mathrm{AF}}\) ସହ ସମାନ୍ତର ଭାବେ ଅଙ୍କିତ ରେଖା \(\overline{\mathrm{FC}}\) କୁ ଓ ବିନ୍ଦୁରେ ଛେଦକରୁ ।
\(\overline{\mathrm{AF}}\), L₂, କୁ H ବିନ୍ଦୁରେ ଛେଦକରୁ ।

ପ୍ରମାଣ : DE = EF (ଦଣ) ଓ \(\stackrel{\leftrightarrow}{\mathrm{BE}}\) || \(\overline{\mathrm{AD}}\) (∵ L₂ || L₁)
⇒ AH = HF,
ଦର୍ମାବ AH = HF ଓ \(\overline{\mathrm{BH}}\) || \(\overline{\mathrm{CF}}\) ⇒ AB = BC
\(\overline{\mathrm{BG}}\) || \(\overline{\mathrm{AF}}\) (ଅକo) ⇒ CG = GF ⇒ CF = 2GF △AFD 6ର H ଓ E ଯଥାକ୍6ମ \(\overline{\mathrm{AF}}\) ଓ \(\overline{\mathrm{DF}}\) ର ମଧ୍ୟଦିଦୁ
⇒ \(\overline{\mathrm{EH}}\) || \(\overline{\mathrm{AD}}\)
∴ ଦଘଣପଷ୍ଟ = CF – AD = 2GF – 2HE
= 2BH – 2HE (∵ GF = BH)
= 2(BH – HE) = 2BE = ଦାମପଖ (ପ୍ରମାଣିତ)

Question 5.

(i) ଭପରିସ୍ଥ ଚିତ୍ର (a) 6ର A – D – B ଏବଂ A – E – C | m∠DAE = 50°, m ∠AED = m∠ABC = 65° | AD = 3 6ସ.ମି., AE : EC = 2 : 1 6ଦୃ6କ, \(\overline{\mathrm{DB}}\) ଓ \(\overline{\mathrm{AB}}\) ର 6ଦିଘ୍ୟ ବ୍ଶଷଯ ଦର |
(ii) ଉପରିସ୍ଥ ବିତ୍ର (b) 6ର \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\), NR = \(\frac { 2 }{ 5 }\) PR ଏବଂ PQ = 10 6ସ. ମି. 6ଦୃ6କ, PM ଓ QM କିଣପ୍ର କର |
(iii) କପଟାସ୍ଟ ଚିତୃ (b) 6ର PM = \(\frac { 1 }{ 2 }\) PQ, NR = 1.2 6ସ.ମି. ଓ \(\overline{\mathbf{M N}} \| \overline{\mathbf{Q R}}\) 6 ଦୁ6 କ, PR ଷ୍ଟିର କର
Solution:
(i) △ ADE 6ର m∠AD = 180° – (50° + 65°) = 65°
∠ADE ≅ ∠ABC ⇒ \(\overline{\mathrm{DE}} \| \overline{\mathrm{BC}}\)

⇒ \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\) = \(\frac { 2 }{ 1 }\)
⇒ \(\frac{3}{DB}\) = \(\frac { 2 }{ 1 }\) ⇒ DB = \(\frac { 3 }{ 2 }\) 6ପ.ମି. = 1.5 6ସ.ମି
AB = AD + DB = 3 6ସ.ମି. + 1.5 6ସ.ମି = 4.5 6ସ.ମି. |

(ii) \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\) (ଦଉ)
\(\frac { PM }{ MQ }\) = \(\frac { PN }{ NR }\) ⇒ \(\frac{P M+M Q}{M Q}\) = \(\frac{\mathrm{PN}+\mathrm{NR}}{\mathrm{NR}}\)

⇒ \(\frac { PQ }{ MQ }\) = \(\frac { PR }{ NR }\) ⇒ \(\frac { 10 }{ MQ }\) = \(\frac{\mathrm{PR}}{\frac{2}{5} \mathrm{PR}}\)
⇒ \(\frac { 10 }{ MQ }\) = \(\frac { 5 }{ 2 }\)
⇒ MQ = \(\frac{10 \times 2}{5}\) 6ସ.ମି. = 4 6ସ.ମି. |
PM = PQ – MQ = 10 6ସ.ମି. – 4 6ସ.ମି. = 6 6ସ.ମି. |
∴ \(\overline{\mathrm{PM}}\) \(\overline{\mathrm{QM}}\) ର ଦେଶ ଯଥାଦୃ6ମ 6 6ସ.ମି. ଓ 4 6ସ.ମି.|

(iii) \(\overline{\mathrm{MN}}\) || \(\overline{\mathrm{QR}}\) , PM = \(\frac { 2 }{ 3 }\) PQ
⇒ MQ = PQ – PM = PQ – \(\frac { 2 }{ 3 }\) PQ = \(\frac { 1 }{ 3 }\) PQ

⇒ \(\frac { 1.2 }{ PR }\) = \(\frac{\frac{1}{3} \mathrm{PQ}}{\mathrm{PQ}}\) ⇒ \(\frac { 1.2 }{ PR }\) = \(\frac { 1 }{ 3 }\)
⇒ PR = 1.2 × 3 6ସ.ମି. = 3.6 ସେ.ମି. |
∴ \(\overline{\mathrm{PR}}\) ର 6ଦଣu = 3.6 ପେ.ପି. |

Question 6.
(i) △ABC 6ର, \(\overline{\mathbf{AB}}\) ଓ \(\overline{\mathbf{AC}}\) ର ମଧତିହୁ ଯଥାକ୍ର6ମ X ଓ Y 6ହ6କ, ଦଶାଥ 6ଯ \(\overline{\mathbf{X Y}} \| \overline{\mathbf{B C}}\) |
(ii) ଏକ ତ୍ରିସ୍ତକର 6ଣାଟିଏ ତାଦୃତ ମଧ୍ୟଝିହୁ 6ଦକ ଅଠ୍ୟ ଏଜ ଚାହପ୍ତତି ଅଜାଡ ସମାତର 6ରଖା, ତୃତୀୟ ବlନୁ ସମକ୍ରିଖଶ୍ନ କ6ର |
(iii) 6ଣାଟିଏ ପମ6କାଣା ତ୍ରିହ୍ନର ଦଶାବ ମଧ୍ୟବିଦୁ ଅଠ୍ୟ ଏବ୍ଲ ଗ୍ଲାସ୍ଥଗି ସମସ୍ତିଖଣ୍ଡକ6ର , ପ୍ରମାଣ କର |
Solution:

Question 7.
△PQR ରେ, \(\overline{\mathbf{PQ}}\) ଓ \(\overline{\mathbf{QR}}\) ବାହୁଦ୍ୱୟର ମଧ୍ୟବିନ୍ଦୁ ଯଥାକ୍ରମେ M ଓ N । \(\overline{\mathbf{PR}}\) ଉପରିସ୍ଥ S ଯେକୌଣସି ଏକ ବିନ୍ଦୁ ହେଲେ, ପ୍ରମାଣ କର ଯେ \(\overline{\mathbf{MN}}\), \(\overline{\mathbf{QS}}\) କୁ ସମର୍ଦ୍ଦିଖଣ୍ଡ କରିବ ।
Solution:
ଦତ : △PQR 6ର \(\overline{\mathbf{PQ}}\) ଓ \(\overline{\mathbf{QR}}\) ର ମଧ୍ୟବିଦୁ ଯଥାଦୃ6ମ M ଓ N | \(\overline{\mathbf{PR}}\) ଭପରିସ୍ଥ ‘S’ ଏକ ବିଦୁ | \(\overline{\mathbf{QS}}\) ଓ \(\overline{\mathbf{MN}}\) ର 6ଛଦଦିଦୁ O |
ସ୍ତାମାଶ୍ୟ : OQ = OS

ପ୍ରମାଣ : \(\frac { QM }{ MP }\) = 1 (∵ QM = MP (ଦର) )
\(\frac { QN }{ NR }\) = 1 (∵ QN = NR (ଦର) )
⇒ \(\frac { QM }{ MP }\) = \(\frac { QN }{ NR }\) ⇒ \(\overline{\mathrm{MN}} \| \overline{\mathrm{PR}}\)
\(\overline{\mathrm{MO}} \| \overline{\mathrm{PS}}\) (∵ \(\overline{\mathrm{MN}} \| \overline{\mathrm{PR}}\) )
⇒ \(\frac { QM }{ MP }\) = \(\frac { QO }{ OS }\) ⇒ 1 = \(\frac { QO }{ OS }\) (∵ \(\frac { QM }{ MP }\) = 1)
⇒ OQ = OS (ପ୍ତମାଣିତ)

Question 8.
ABCD ଣ୍ଠାପିଲିଷ୍ଟମ6ର \(\overline{\mathbf{AB}} \| \overline{\mathbf{CD}}\) | ଲଣ \(\overline{\mathbf{AC}}\) ଓ \(\overline{\mathbf{BD}}\) ର 6ବ୍ଳଦଦିନ୍ଦି P 6ଦୃ6କ , ପ୍ତମାଶା କର 6ମ
(i) AP : PC = BP : PD
(ii) CP : AC = DP : BD |
Solution:

Question 9.
ABCD ଗ୍ରାପିଦିକ୍ଷ୍ମମ6ର \(\overline{\mathbf{AB}}\) || \(\overline{\mathbf{DC}}\) ଏର \(\overline{\mathbf{AD}}\) ର ମଧ୍ୟତିଦୁ P | \(\overline{\mathbf{AB}}\) ସଦୃ ସାପବ୍ଦର ଉ6ର ଅବିର \(\overleftrightarrow{\mathbf{P Q}}\) \(\overline{\mathbf{BC}}\) କ Q ଦିହି6ର 6ଛଦନ6କ , ପ୍ରମାଣ ଦ୍ଦର 6ଯ Q 6ଦୃରଚି \(\overline{\mathbf{BC}}\) ର ମଧ୍ୟନିନ୍ଦୁ |
Solution:

Question 10.
ABCD ଉପରୋକ୍ତ \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{BC}}\) , \(\overline{\mathrm{CD}}\) ,ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ସର୍ବସମ P, Q, R ଓ S |
(a) ପ୍ରମାଣ କର ଯେ PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର ।
(b) ଉପରୋକ୍ତ ଚତୁର୍ଭୁଜ ABCD ର କର୍ଣ୍ଣଦ୍ଵୟ ପରସ୍ପର ପ୍ରତି ଲମ୍ବ ହେଲେ, ପ୍ରମାଣକର ଯେ PORS ଏକ ଆୟତ ଚିତ୍ର ।
Solution:
(a) ଦତ୍ତ : ABCD ଉପରୋକ୍ତ \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{BC}}\) , \(\overline{\mathrm{CD}}\) , ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ସର୍ବସମ P, Q, R ଓ S |
ପ୍ରାମାଣ୍ୟ : PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର ।
ଅଙ୍କନ : \(\overline{\mathrm{BD}}\) ଅଙ୍କନ କରାଯାଉ ।
ପ୍ରମାଣ : △ABD ରେ \(\overline{\mathrm{AB}}\) ର ମଧ୍ୟବିନ୍ଦୁ P ଓ \(\overline{\mathrm{AD}}\) ର ମଧ୍ୟବିନ୍ଦୁ S ।
\(\frac { AP }{ BP }\) = 1 (∵ AP = BP (ଦତ୍ତ))
ସେହିପରି ଧୂ) \(\frac { AS }{ SD }\) = 1 (∵ AS = SD (ଦତ୍ତ))

∴\(\frac { AP }{ BP }\) = \(\frac { AS }{ SD }\) ⇒ \(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\)
ସେହିପରି ପ୍ରମାଣ କରାଯାଇ ପାରିବ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BD}}\) |
\(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\) ଓ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{BD}}\)
\(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{QR}}\)
\(\overline{\mathrm{AC}}\) ଅଙ୍କନ କରି ସେହିପରି ପ୍ରମାଣ କରାଯାଇ ପାରେ ଯେ \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{SR}}\) |
∴ PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର |

(b) ଦତ୍ତ : ABCD ଚତୁର୍ଭୁଜରେ \(\overline{\mathrm{AC}}\) ⊥ \(\overline{\mathrm{BD}}\) | \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CD}}\) ଓ \(\overline{\mathrm{DA}}\) ର ମଧ୍ୟବିନ୍ଦୁ ଯଥାକ୍ରମେ P, Q, R, S |
ପ୍ରାମାଣ୍ୟ : PORS ଏକ ଆୟତଚିତ୍ର ।
ପ୍ରମାଣ : ମନେକର \(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{BD}}\) ଛେଦବିନ୍ଦୁ ( ଏବଂ \(\overline{\mathrm{PS}}\) ଓ \(\overline{\mathrm{AC}}\) ର ଛେଦବିନ୍ଦୁ X ।

(a) ର ଅନୁରୂପ ପ୍ରମାଣ ଦ୍ଵାରା PORS ଏକ ସାମାନ୍ତରିକ ଚିତ୍ର 6ଚିତ୍ର ।
m∠AOB = 90° ⇒ m∠PXO = 90°
(∵ \(\overline{\mathrm{PS}}\) || \(\overline{\mathrm{BD}}\))
m∠XPXQ = 90° (∵ \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\))
∴ PQRS ଏକ ସାମାନ୍ତରିକ |

Question 11.
ନିମ୍ନ ଚିତ୍ରରେ △ABC ର \(\overline{\mathrm{BA}}\) ବାହୁ ସହ \(\overline{\mathrm{CM}}\) ସମାନ୍ତର, \(\overline{\mathrm{AB}}\) ର ପ୍ରାମାବିନ୍ଦୁ P | \(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\), \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{CM}}\); ସାମାନ୍ତରିକ ମେ
\(\overline{\mathrm{PR}}\) || \(\overline{\mathrm{AM}}\) |
Solution:
ଦତ୍ତ : △ABC ରେ \(\overline{\mathrm{BA}}\) || \(\overline{\mathrm{CM}}\), AP = BP,
\(\overline{\mathrm{PQ}}\) || \(\overline{\mathrm{AC}}\) ଓ \(\overline{\mathrm{QR}}\) || \(\overline{\mathrm{CM}}\)
ପ୍ରାମାଣ୍ୟ : \(\overline{\mathrm{PR}}\)||\(\overline{\mathrm{AM}}\)
ପ୍ରମାଣ : △ABC ରେ \(\overline{\mathrm{PQ}}\) ||\(\overline{\mathrm{AC}}\) (ଦତ୍ତ) ।

⇒ \(\frac { BP }{ AP }\) = \(\frac { BQ }{ QC }\)
△BCM ରେ \(\overline{\mathrm{QR}}\) ||\(\overline{\mathrm{CM}}\) (ଦତ୍ତ) ।
⇒ \(\frac { BQ }{ QC }\) = \(\frac { BR }{ RM }\) ……(i)
∴ (i) ଓ (ii)ରୁ \(\frac { BP }{ AP }\) = \(\frac { BR }{ RM }\)
△BCM ରେ \(\frac { BP }{ AP }\) = \(\frac { BR }{ RM }\) ⇒ \(\overline{\mathrm{PR}}\) ||\(\overline{\mathrm{AM}}\) (ପ୍ରମାଣିତ)
ତ୍ରିଭୁଜର କୋଣର ସମଦ୍ବିଖଣ୍ଡକ ସମ୍ବନ୍ଧୀୟ ଆଲୋଚନା

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Ex 12(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Exercise 12(a)

Question 1.
Fill in the blanks by choosing the correct answer from the given alternatives :
(a) The center of the circle x² + y² + 2xy – 6y + 1 = 0 is _____________. [(2, -6), (-2, 6), (-1, 3), (1, -3)]
Solution:
(-1, 3)

(b) The equation 2x² – ky² – 6x + 4y – 1 = 0 represents a circle if k = ____________. [2, -2, 0, 1]
Solution:
-2

(c) The point (-3, 4) lies ______________ the circle x² + y² = 16 [outside, inside, on]
Solution:
Outside

(d) The line y = x + k touches the circle x² + y² = 16 if k = _______________. [±2√2, ±4√2, ±8√2, ±16√2]
Solution:
±4√2

(e) The radius of the circle x² + y² – 2x + 4y + 1 = 0 is _______________. [1, 2, 4, √19]
Solution:
2

Question 2.
State (with reasons), which of the following is true or false :
(a) Every second-degree equation in x and y represents a circle.
Solution:
Every 2nd-degree equation in x and y represents a circle if the coefficients of x and y are equal and the equation does not contain xy term (False)

(b) The circle (x – 1)² + (y – 1)² = 1 passes through origin.
Solution:
(0 – 1)² + (0 – 1)² = 1 + 1 = 2 ≠ 1.
So the circle does not pass through the origin. (False)

(c) The line y = 0 is a tangent to the circle (x + 1)² + (y – 2)²  = 1.
Solution:
The line y = 0 is a tangent to the circle centre at (-1, 2) and the radius is 1. (True)
∴ The distance of the centre from the line y = 0 is 1 which is equal to its radius.

(d) The radical axis of two circles always passes through the centre of one of the circles,
Solution:
As radical axis is the common chord of the circles, which should not pass through the centre of one of the circles. (False)

(e) The circle x² + (y – 3)² = 4 and (x – 4)² + y² = 9 touch each other.
Solution:
The distance between the centres is \(\sqrt{(0-4)^2+(3-0)^2}\) = 5 which is equal to the sum of the radii. (True)

Question 3.
Find the equation of circles determined by the following conditions.
(a) The centre at (1, 4) and passing through (-2, 1).
Solution:

(b) The centre at (-2, 3) and passing through origin.
Solution:
Centre at (-2, 3) and circle passes through origin.
∴ Radius of the circle = \(\sqrt{(-2)^2+3^2}=\sqrt{13}\)
∴ Equation of the circle is (x – h)² + (y – k)² = a²
or, (x + 2)² + (y – 3)² = 13

(c) The centre at (3, 2) and a circle is tangent to x – axis.
Solution:

(d) The centre at (-1, 4) and circle is tangent to y – axis.
Solution:

(e) The ends of diameter are (-5, 3) and (7, 5).
Solution:
The endpoints of the diameter of the circle are (-5, 3) and (7, 5).
∴ Equ. of the circle is
(x – h)² + (y – k)² = a²
(x- x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
or, (x + 5)(x – 7) + (y – 3)(y – 5) = 0
or, x² – 7x + 5x – 35 + y² – 5y – 3y + 15 = 0
or, x² + y² – 2x – 8y – 20 = 0

(f) The radius is 5 and circle is tangent to both axes.
Solution:
As the circle is tangent to both axes, we have its centre at (5, 5).
∴ Equation of the circle is
or, (x ± 5)² + (y ± 5)² = 25

(g) The centre is on the x-axis and the circle passes through the origin and the point (4, 2).
Solution:

∴ \(\sqrt{(4-a)^2+4}\) = a
or, (4 – a)² + 4 = a²
or, 16 + a² – 8a + 4 = a²
or, 8a = 20 or, a = \(\frac{20}{8}=\frac{5}{2}\)
∴ Equation of the circle is
(x – h)² + (y – k)² = a²
or, (x – \(\frac{5}{2}\))² + (y – 0)² = (\(\frac{5}{2}\))²
or, x² + \(\frac{25}{4}\) – 5x + y² = \(\frac{25}{4}\)
or, x² + y² – 5x = 0

(h) The centre is on the line 8x + 5y = 0 and the circle passes through the points (2, 1) and (3, 5).
Solution:
Let the equation of the circle be x² + 2gx + y² + 2fy + c = 0
∴ Its centre at (- g, -f). As the centre lies on the line 8x + 5y = 0
We have -8g – 5f = 0      …..(1)
Again, as the circle passes through points (2, 1) and (3, 5)
We have
4 + 4g + 1 + 2f + c = 0  …..(2)
and 9 + 6g + 25 + 10f + c = 0   …..(3)
Now from (1), we have g = \(\frac{-5 f}{8}\)
From equation (2), 4g + 2f + c + 5 = 0
or, 4 \(\frac{-5 f}{8}\) + 2f + c + 5 = 0
or, -5f + 4f + 2c + 10 = 0
or, f = 2c + 10    …..(4)
(2) 6g + 10f + c + 34 = 0
or, 6\(\frac{-5 f}{8}\) + 10f + c + 34 = 0
or, -15f + 40f + 4c + 136 = 0
or, 25f = -4c – 136
or, f = \(\frac{-4 c-136}{25}\)
∴ 2c + 10 = \(\frac{-4 c-136}{25}\)
or, 25 (c + 5) = -2c – 68
or, 25c + 2c = -68 – 125
or, 27c = -193 or, c = \(\frac{-193}{27}\)
∴ f = 2C + 10 = 2(\(\frac{-193}{27}\)) + 10
= \(\frac{-386+270}{27}=\frac{-116}{27}\)
∴ g = \(\frac{-5 f}{8}=\frac{-5}{8} \times\left(\frac{-116}{27}\right)=\frac{145}{54}\)
Eqn. of the circle is x² + y² + 2 × \(\frac{145}{54}\) x + 2 \(\frac{-116}{27}\) y + \(\frac{-193}{27}\) = 0
or, 27x² + 27y² + 145x – 232y – 193 = 0

(i) The centre is on the line 2x + y – 3 = 0 and the circle passes through the points (5, 1) and (2, -3).
Solution:
Let the eqn. of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through (5, 1) and (2, -3),
we have 25 + 1 + 10g + 2f + c = 0 …(1)
and 4 + 9 + 4g – 6f + c = 0    …..(2)
Again as the centre lies on the line 2x + y – 3 = 0,
we have- 2g – f – 3 = 0 or, f= -2g – 3
∴ From equation (1)
10g + 2 (-2g – 3) + c + 26 = 0
or, 10g – 4g – 6 = -c – 26
or, 6g = -c – 20
or, g = \(\frac{-c-20}{6}\)
∴ From equation (2)
4g – 6 (-2g – 3) + c + 13 = 0
or, 4g + 12g + 18 + c + 13 = 0
or, 16g = -c – 31
or, g = \(\frac{-c-31}{16}\)
∴ \(\frac{-c-20}{6}=\frac{-c-31}{16}\)
or, -8c – 160 = -3c – 93
or, 5c = -160 + 93 = -67
or, c = –\(\frac{67}{5}\)
∴ g = \(\frac{-c-20}{6}=\frac{\frac{67}{5}-20}{6}=\frac{67-100}{5 \times 6}\)
= \(\frac{-33}{5 \times 6}=\frac{-11}{10}\)
∴ f = -2g – 3 = (-2)\(\left(\frac{-11}{10}\right)\)
= \(\frac{11-15}{5}=\frac{-4}{5}\)
∴ Eqn. of the circle is x² + y² + 2 (\(\frac{-11}{10}\))x + 2(\(\frac{-4}{5}\))y – \(\frac{67}{5}\) = 0
or, 5x² + 5y² – 11x – 8y – 67 = 0

(j) The circle is tangent to the line x + 2y – 9 = 0 at (5, 2) and also tangent to the line 2x – 3y – 7 = 0 at (2, -1).
Solution:

(k) The circle touches the axis of x at (3, 0) and also touches the line 3y – 4x = 12.
Solution:
Let the centre be at (3, k)
Radius = k

or, 3k – 24 = ±5k or, 2k = -24
or, k = -12
Also k = 3
∴ Equation of the Circle is
(x – 3)² + (y – k)² = k²
or, x² + 9 – 6x + y² + k² – 2ky = k²
or, x² + y2 – 6x – 2 (-12)y = 0
or, x² + y² – 6x + 24y + 9 = 0
and x² + y² – 6x – 6y + 9 = 0

(l) Circle is tangent to x – axis and passes through (1, -2) and (3, -4).
Solution:
Let the centre be at (h, k).
So the radius is k.

∴ Equation of the circle is (x – h)² + (y – k)² = k²
or, (x + 5)² + (y + 10)² = 100 and (x – 3)² + (y + 2)² = 4

(m) Circle passes through origin and cuts of intercepts a and b from the axes.
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0

(n) Circle touches the axis of x at a distance of 3 from the origin and intercepts a distance of 6 on the y-axis.
Solution:
Let the centre be at (3, k).
So the radius is k.
∴ Equation of the circle is (x – 3)² + (y – k)² = k²
or, x² + 9 – 6x + y² + k² – 2ky = k²
or, x² + y² – 6x – 2xy + 9 = 0

∴ |y₂ – y₁| = 2\(\sqrt{k^2-9}\) = 6
or, \(\sqrt{k^2-9}\) = 3
or, k² = 18, or, k = ±3√2
∴ Equation of the circle is x² – y² – 6x ± 6y√2 + 9 = 0

Question 4.
Find the centre and radius of the following circles:
(a) x² + y² + 6xy – 4y – 12 = 0
Solution:
x² + y² + 6xy – 4y – 12 = 0
∴ 2g = 6, 2f = – 4, c = -12
∴ 8 = 3, f = -2
Centre of (-g, -f) = (-3, 2) and radius = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{9+4+12}\) = 5

(b) ax² + ay² + 2gx + 2fy + k = 0
Solution:
ax² + ay² + 2gx + 2fy + k = 0
or, x² + y² + \(\frac{2 g}{a}\)x + \(\frac{2 f}{a}\)y + \(\frac{k}{a}\) = 0
∴ Centre of \(\left(\frac{-g}{a}, \frac{-f}{a}\right)\)
and radius = \(\sqrt{\frac{g^2}{a^2}+\frac{f^2}{a^2}-\frac{k}{a}}=\sqrt{\frac{g^2+f^2-a k}{a}}\)

(c) 4x² + 4y² – 4x + 12y – 15 = 0
Solution:
4x² + 4y² – 4x + 12y – 15 = 0
or, x² + y² – 4 + 3y  – \(\frac{15}{4}\) = 0
∴ 2g = -1, 2f = 3, c = \(\frac{15}{4}\)
∴ g = – \(\frac{1}{2}\), f = \(\frac{3}{2}\)
∴ Centre at (-g, -f) = (\(\frac{1}{2}\), \(\frac{-3}{2}\)) and radius \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{\frac{1}{4}+\frac{9}{4}+\frac{15}{4}}=\frac{5}{2}\)

(d) a(x² + y²) – bx – cy = 0
Solution:
a(x² + y²) – bx – cy = 0
or, x² + y² – \(\frac{b x}{a}\) – \(\frac{c y}{a}\) = 0

Question 5.
Obtain the equation of circles passing through the following points and determine the coordinates of the centre and radius of the circle in each case:
(a) the points (3, 4) (4, -3) and (-3, 4).
Solution:
Let the centre be at (h, k)

(b) the points (2, 3), (6, 1) and (4, -6).
Solution:
Let the centre be at (h, k).

we have \(|\overline{\mathrm{PC}}|=|\overline{\mathrm{QC}}|=|\overline{\mathrm{RC}}|\)
∴ (h – 4)² + (k + 6)² = (h – 6)² + (k – 1)² and (h – 4)² + (k + 6)² = (h – 2)² + (k – 3)²
∴ h² + 16 – 8h + k² + 36 + 12k
= h² + 36 – 12h + k² + 1 – 2k
and h² + 16 – 8h + k² + 36 + 12k
= h² + 4 – 4h + k² + 9 – 6k
or, 14k = -4h – 15 and 18k = 4h – 39
or, k = \(\frac{-4 h-15}{14}\) and k = \(\frac{4 h-39}{18}\)

(c) the points (a, 0), (-a, 0) and (0, b).
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0. As it passes through the points (a, 0), (-a, 0) and (0, b). We have
a² – 2ga + c = 0   …..(1)
a² + 2ga + c = 0   …..(2)

(d) the points (-3, 1), (5, -3) and (-3, 4).
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through the points. we have (-3, 1), (5, -3) and (3, 4).
We have 9 + 1 – 6g + 2f + c = 0    …..(1)
25 + 9 + 10g – 6df + c = 0      …(2)
9 + 16 – 6g + 8f + c = 0     …..(3)

Question 6.
Find the equation of the circles circumscribing the triangles formed by the lines given below :
(a) the lines x = 0, y = x, 2x + 3y = 10
Solution:

∴ The coordinates. C are (0, 0) of
Lastly, solving \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)
we have y = x, 2x + 3y = 1 0
we have 5x = 10
or, x = 2 and y = 2.
∴ The coordinates of A are (2, 2).
∴ The circle passes through the points (2, 2), (0, \(\frac{10}{3}\)) and (0, 0)
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through the points A, B, C we have c = 0, 4 + 4 + 4g + 4f = 0,
\(\frac{100}{9}\) + 2. f. \(\frac{10}{3}\) + 0 = 0
∴ f = \(\frac{-100}{9} / \frac{20}{3}=\frac{-5}{3}\)
and g = \(\frac{-4 f-8}{4}=\frac{-4\left(\frac{-5}{3}\right)-8}{4}\)
= \(\frac{20-24}{3 \times 4}=\frac{-1}{3}\)
∴ Equation of the circle is  x² + y² + 2(\(\frac{-1}{3}\))x + 2 \(\frac{-5}{3}\)y + 0 = 0
or, 3(x² +  y²) – 2x – 10y = 0

(b) The lines x = 0, 4x + 5y = 35, 4y = 3x + 25
Solution:



or, 4x² + 4y² – 24x – 53y + 175 = 0

(c) The lines x = 0, y = 0, 3x + 4y – 12 = 0
Solution:
The coordinates of A, B and C are (4, 0), (0, 3) and (0, 0).
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0

(d) The lines y = x, y =2 and y = 3x + 2
Solution:

(e) the lines x + y = 6, 2x + y = 4 and x + 2y = 5
Solution:

Question 7.
Find the coordinates of the points where the circle x² + y² – 7x – 8y + 12 = 0 meets the coordinates axes and hence find the intercepts on the axes. [Hint: If a circle intersects a line at points A and B, then the length AB is its intercepts on line L]
Solution:
x² + y² – 7x – 8y + 12 = 0
Putting x = 0, we have y² – 8y + 12 = 0 or, (y – 6) (y – 2) = 0, or, y = 6, 2.
∴ The circle meets the Y-axis at (0, 6) and (0, 2) and its Y-intercept is 6 – 2 = 4.
Again putting y = 0,
we have x² – 7x + 12 = 0
or, (x – 4)(x – 3) = 0 or, x = 4, x = 3.
∴ The circle meets the X-axis at (4, 0) and (3, 0) and its x-intercept is 4 – 3 = 1.

Question 8.
Find the equation of the circle passing through the point (1, -2) and having its centre at the point of intersection of lines 2x – y + 3 = 0 and x + 2y – 1 =0
Solution:

Question 9.
Find the equation of the circle whose ends of a diameter are the points of intersections of the lines and x + y – 1 = 0, 4x + 3y + 1 = 0 and 4x +y + 3 = 0, x – 2y +3 = 0.
Solution:
Solving x + y – 1 = 0, 4x + y + 3 = 0

∴ The endpoints of the diameter are (-4, 5) and (-1, 1).
∴ Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
or, (x + 4) (x + 1) + (y – 5) (y – 1) = 0
or, x² + x + 4x + 4 + y² – y – 5y + 5 = 0
or, x² + y² + 5x – 6y + 9 = 0.

Question 10.
Find the equation of the circle inscribed inside the triangle formed by the line \(\frac{x}{4}+\frac{y}{3}\) = 1 and the coordinate axes.
Solution:
The circle is inscribed in the triangle formed by x = 0, y = 0 and \(\frac{x}{4}+\frac{y}{3}\) = 1
∴ If (h, k) is the centre and r is the radius of the circle then h = k = r.
The perpendicular distance of the centre (h, h) from the line 3x + 4y = 12 is the radius.
⇒ \(\left|\frac{3 h+4 h-12}{5}\right|\) = h
⇒ 7h – 12 = ±5h
⇒ 2h = 12 or 2h = 12
⇒ h = 6 or h = 1
But h can not be 6 thus the circle has equation (x – 1)² + (y – 1)² = 1
⇒ x² + y² – 2x – 2y + 1 =0

Question 11.
(a) Find the equation of the circle with its centre at (3, 2) and which touches to the line x + 2y – 4 = 0.
Solution:

(b) The line 3x + 4y + 30 = 0 is a tangent to the circle whose centre is at (\(-\frac{12}{5},-\frac{16}{5}\)). Find the equation of the circle.
Solution:

(c) Prove that the points (9, 7), and (11, 3) lie on a circle with centre at origin. Find the equation of the circle.
Solution:

(d) Find the equation of the circle which touches the line x = 0, x = a and 3x + 4y + 5a = 0.
Solution:

(e) If a circle touches the co-ordinate axes and also touches the straight line \(\frac{x}{a}+\frac{y}{b}\) = 1 and has its centre in the 1st quadrant, And its equation.
Solution:
Let the centre be at (k, k) and the radius is k.

Question 12.
ABCD is a square of side ‘a’ If AB and AD are taken as co-ordinate axes, prove that the equation of the circle circumscribing the square is
x² + y² = a(x + y)
Solution:

or, x² + \(\frac{a^2}{4}\) – ax + y² + \(\frac{a^2}{4}\) – ay = \(\frac{a^2}{2}\)
or, x² + y² – ax – ay = 0
or, x² + y² = a(x + y)

Question 13.
(a) Find the equation of the tangent and normal to the circle x² + y² = 25 at the point (3, -4).
Solution:
Equation of the tangent to the circle x² + y² = 25 at the point (3, -4) is
xx₁ + yy₁ = a²
3x – 4y = 25
Equation of the normal is x₁y = xy₁
or, 3y = -4x or, 4x + 3y = 0

(b) Find the equation of the tangent and normal, to the circle, x² + y² – 3x + 4y – 31 = 0 at the point (-2, 3).
Solution:
Equation of the tangent of the circle x² + y² – 3x + 4y – 31 = 0 at the point (-2, 3) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, -2x + 3y – \(\frac{3}{2}\) (x – 2) + 2(y + 3) – 31 = 0
or, – 4x + 6y – 3x + 6 + 4y + 12 – 62 = 0
or, -7x + 10y – 44 = 0
or, 7x – 10y + 44 = 0
Equation of the normal is x(f + y₁) – y(g + x₁) fx₁ + gy₁ = 0
or, x(2 + 3) – y(\(-\frac{3}{2}\) – 2) – 2(-2) – \(\frac{3}{2}\) × 3 = 0
or, 5x + \(\frac{7y}{2}\) + 4 – \(\frac{9}{2}\) = 0
or, 10x + 7y – 1 = 0

(c) Find the equation of the tangents to the circle x² + y² + 4x – 6y – 16 = 0 at the point where it meets the y – axis.
Solution:
Putting x = 0 in the circle equation, we have
y² – 6y – 16 = 0
or, y² – 8y + 2y – 16 = 0
or, y(y – 8) + 2(y – 8) = 0
or, (y – 8)(y + 2) = 0
y = 8 or, -2
The circle meets y – axis at (0, 8) and (0, -2).
Eqn. of the tangents are
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, 0 + 8y + 2 (x + 0) – 3(y + 8) – 16 = 0
or, 8y + 2x – 3y – 24 – 16 = 0
or, 2x + 5y = 40 and
x × 0 – 2y + 2 (x + 0) – 3 (y – 2) – 16 = 0
or, -2y + 2x – 3y + 6 – 16 = 0
or, 2x – 5y – 10 = 0

(d) Find the condition under which the tangents at (x₁, y₁) and (x₂, y₂) to the circle x² + y² + 2gx + 2fy + c = 0 are perpendicular.
Solution:
Equation of tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at the point (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, (g + x₁)x + y(f + y₁) + gx₁ + fy₁ + c = 0
Again equation of the tangent to the circle at (x₂, y₂) is
x(g + x₂) + y(f + y₂) + gx₂ + fy₂ + c = 0
As the tangent (1) and (2) are perpendicular, we have the product of their slopes is -1.
∴ \(\frac{g+x_1}{f+y_1} \times \frac{g+x_2}{f+y_1}\) = -1
or, (g + x₁)(g + x₂) = -(f + y₁)(f + y₂)
or, (g + x₁)(g + x₂) + (f + y₁)(f + y₂) = 0

(e) Calculate the radii and distance between the centres of the circles, whose equations are, x² + y² – 16x – 10y + 8 = 0; x² + y² + 6x – 4y – 36 = 0. Hence or otherwise prove that the tangents drawn to the circles at their points of intersection are perpendicular.
Solution:
x² + y² – 16x – 10y + 8 = 0;
x² + y² + 6x – 4y – 36 = 0.
g₁ = -8, f₁ = -5, c₁ = 8,
g₂ = 3, f₂ = -2, c₂ = -36
The centres are (-g₁, -f₁) and (-g₂, -f₂)

Question 14.
(a) Find the equation of the tangents to the circle x² + y² = 9 perpendiculars to the line x – y – 1 = 0
Solution:

(b) Find the equation of the tangent to the circle x² + y² – 2x – 4y = 40, parallel to the line 3x – 4y = 1.
Solution:


(c) Show that the line x – 7y + 5 = 0 a tangent to the circle x² + y² – 5x + 5y = 0. Find the point of contact. Find also the equation of tangent parallel to the given line.

Solution:
we have the line is x – 7y + 5 = 0
or, y = \(\frac{x+5}{7}\)
Now putting the value of y in the circle, we have x² + y² – 5x + 5y = 0
or, x² + (\(\frac{x+5}{7}\))² – 5x + 5 \(\frac{x+5}{7}\) = 0
or, 49x² + x² + 25 + 10x – 245x + 35x + 175 = 0
or, 50x² – 200x + 200 = 0
or, x² – 4x + 4 = 0
∴ a = 1, b = -4, c = 4
∴ b² – 4ac = (-4)² – 4 × 1 × 4
= 16 – 16 = 0
∴ The line x – 7y + 5 = 0

x – 7y – 45 and x – 7y + 5 = 0

(d) Prove that the line ax + by + c = 0 will be the tangent to the circle x² + y² = r² if r²(a² + b²) = c².
Solution:
We know that a line is a tangent to the circle if the distance of the line from the centre is equal to the radius.
Now the circle is x² + y² = r²
⇒ Centre is at (0, 0) and radius r. The distance of (0, 0) from ax + by + c = 0 is

(e) Prove that the line 2x + y = 1 tangent to the circle x² + y² + 6x – 4y + 8 = 0.
Solution:

(f) If the line 4y – 3x = k is a tangent to the circle x² + y² + 10x – 6y + 9 = 0 find ‘k’. Also, find the coordinates of the point of contact.
Solution:
Center of the circle is (-5, 3) and the radius is \(\sqrt{25+9-9}\) = 5
Distance of the centre from the line 4y – 3x – k = 0

Question 15.
(a) Find the length of the tangent, drawn to the circle x² + y² + 10x – 6y + 8 = 0 from the centre of the circle x² + y² + 4x = 0.
Solution:
Center of the circle x² + y² + 4x = 0 is (2, 0)
∴ Length of the tangent drawn from the point (2, 0) to the circle x² + y² + 10x – 6y + 8 = 0
\(\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 g \mathrm{~g}_1+2 \mathrm{fy}_1+\mathrm{c}}\)
= \(\sqrt{4+0+10 \times 2+0+8}=\sqrt{32}=4 \sqrt{2}\)

(b) Find the length of the tangent drawn from the point (2, -1) to the circle x² + y² + 6x + 10y + 18 = 0
Solution:
Length of the tangent drawn from the point (2, -1) to the circle x² + y² + 6x + 10y + 18 = 0 is
\(\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\)
= \(\sqrt{4+1+(-6) \times 2+10(-1)+18}\)
= \(\sqrt{5-12-10+18}\) = 1

(c) Find the length of the tangent drawn from the point (4, 7) to the circle x² + y² = 15.
Solution:
Length of the tangent drawn from the point (4, 7) to the circle x² + y² = 15 is \(\sqrt{16+49-15}\) = √50 = 5√2

Question 16.
(a) Prove that the circle given by the equations x² + y² + 2x – 8y + 8 = 0 and x² + y² + 10x – 2y + 22 = 0 touches each other externally. Find also the point of contact
Solution:
x² + y² + 2x – 8y + 8 = 0
g₁ = 1, f₁ = -4, c₁ = 8
Hence centre = c₁(-g₁, -f₁) = c₁(-1, 4)
Radius = r₁ = \(\sqrt{1+16-8}\) = 3
Again x² + y² + 10x – 2y + 22 = 0
g₂ = 5, f₂ = -1, c₂ = 22
Centre c₂(-g₂, -f₂) = c₂(-5, 1)
Radius r₂ = \(\sqrt{25+1-22}\) = 2
Now

(b) Prove that the circle is given by the equations x² + y² = 4 and x² + y² + 6x + 8y – 24 = 0, touch each other and find the equation of the common tangent.
Solution:
x² + y² = 4,
x² + y² + 6x + 8y – 24 = 0
Their centres are (0, 0) and (-3, -4) and radii are 2 and \(\sqrt{9+16+24}\) = 7
∴ Distance between the centres is \(\sqrt{(-3)^2+(-4)^2}\) = 5, which is equal to the difference between the radii.
∴ The circles touch each other internally.
∴ Equation of the common tangent is S₁ – S₂ = 0
or, (x² + y² + 6x + 8y – 24) – (x² + y² – 4) = 0
or, 6x + 8y – 20 = 0
or, 3x + 4y = 10

(c) Prove that the two circle x² + y² + 2by + c² = 0 and x² + y² + 2ax + c² = 0,  will touch each other \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\).
Solution:
x² + y² + 2by + c² = 0,
x² + y² + 2ax + c² = 0
g₁ = 0, f₁ = b, c₁ = c².
g₂ = a, f₂ = 0, c₂ = c².
The centres of the circle are (0, -b) and (-a, 0) and radii are \(\sqrt{b^2-c^2}\) and \(\sqrt{a^2-c^2}\). As the circles touch each other, we have the distance between the centres is equal to the sum of the radii.

(d) Prove that the circles given by x² + y² + 2ax + 2by + c = 0, and x² + y² + 2bx + 2ay + 2c = 0, touch each other, if (a + b) = 2c.
Solution:
x² + y² + 2ax + 2by + c = 0
x² + y² + 2bx + 2ay + 2c = 0,
The centre of the circle is (-a, -b) and (-b, -a). the radii of the circle are \(\sqrt{a^2+b^2-c}\) and \(\sqrt{b^2+a^2-c}\). As the circles touch each other we have, the distance between the centres is equal to the sum of the radii.

Question 17.
Find the equation of the circle through the point of intersection of circles x² + y² – 6x = 0 and x² + y² + 4y – 1 = 0 and the point (-1, 1).
Solution:
Let the equation of the circle be (x² + y² – 6x) + λ(x² + y² + 4y – 1) = 0
As it passes through the point (-1, 1),
we have (1 + 1 + 6) + λ(1 + 1 + 4 – 1) = 0
or, 8 + 5λ = 0 or, λ = \(\frac{-8}{5}\)
∴ Equation of the circle is (x² + y² – 6x) – \(\frac{8}{5}\) (x² + y² + 4y – 1) = 0
or, 5x² + 5y² – 30x – 8x² – 8y² – 32y + 8 = 0
or, 3x² + 3y² + 30x + 32y – 8 = 0

Question 18.
Find the equation of the circle passing through the intersection of the circles, x² + y² – 2ax = 0 and x² + y² – 2by = 0 and having the centre of the line \(\frac{x}{a}-\frac{y}{b}\) = 2
Solution:

Question 19.
Find the radical axis of the circles x² + y² – 6x – 8y – 3 = 0 and 2x² + 2y² + 4x – 8y = 0
Solution:
x² + y² – 6x – 8y – 3 = 0
2x² + 2y² + 4x – 8y = 0
x² + y² – 6x – 8y – 3 = 0
x² + y² + 2x – 4y = 0
∴ The equation of the radical axis is S₁ – S₂ = 0
or, (x² – y²– 6x – 8y – 3) – (x² + y² + 2x – 4y) = 0
or, -6x – 8y- 3 – 2x + 4y = 0
or, -8x – 4y – 3 = 0
or, 8x + 4y + 3 = 0

Question 20.
Find the radical axes of the circle x² + y² – 6x + 8y – 12 = 0  and x² + y² + 6x – 8y + 12 = 0 Prove that the radical axis is perpendicular to the line joining the centres of the two circles.
Solution:
Equation of the radical axes of the circle x² + y² – 6x + 8y – 12 = 0  and x² + y² + 6x – 8y + 12 = 0
(x² + y² – 6x + 8y – 12) – (x² + y² + 6x – 8y + 12) = 0
or, -12x + 16y – 24 = 0
or, 3x – 4y + 6 = 0
Again, slope of the radical axis is \(\frac{3}{4}\) = m₁ (say)
Centres of the circles are (3, -4) and (-3, 4).
Slope of the line joining the centres is \(\frac{4+4}{-3-3}=\frac{8}{-6}=-\frac{4}{3}\) = m₂ (say)
m₁. m₂ = \(\frac{3}{4}\left(-\frac{4}{3}\right)\) = -1
∴ The radical axis is perpendicular to the line joining centres of the circles. (Proved)

Question 21.
If the centre of one circle lies on or inside another, prove that the circles cannot be orthogonal.
Solution:
The orthogonality condition for two circles.
x² + y² + 2g₁x + 2f₁y + C₁ = 0   …..(1)
and x² + y² + 2g₂x + 2f₂y + C₂ = 0    …..(2)
is 2(g₁g₂ + f₁f₂) – C₁ – C₂ = 0
Let us consider two circles
Case-1. Let the centre of (2) which is C (-g₂, -f₂) lies on the circle (1). Hence it satisfies the equation (i)
i.e., g₂² + f₂² – 2g₁g₂ – 2f₁f₂ + C₂ = 0
⇒ 2g₁g₂ + 2f₁f₂ – C₁ – C₂ = g₂² + f₂² – C₂
Its right-hand side is the square of the radius of 2nd circle which can not be equal
to zero i.e., 2(g₁g₂ + f₁f₂) – C₁ – C₂ ≠ 0
Hence circles are not orthogonal.
Case-2. Let the centre of (2) which is (-g2, -f2) lies inside the circle (1).
Distance between their centres < radius of the first circle.
i,e. \(\sqrt{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2}<\sqrt{g_1^2+f_1^2-C_1}\)
⇒ g₁² – 2g₁g₂ + f₁² + f₂² + 2f₁f₂ < g₁² + f₁² – C₁
⇒ 2g₁g₂ – 2f₁f₂ – C₁ – C₂ > g₂² + f₂² – C₂
= square of the radius of 2^nd circle. Hence greater than 0.
⇒ 2(g₁g₂ + f₁f₂) – C₁ – C₂ > 0
So two circles are not orthogonal. By case -1 and case -2 we conclude that if the centres of one circle lie on or inside another, then circles cannot be orthogonal.

Question 22.
If a circle S intersects circles S₁ and S₂ orthogonally. Prove that the centre of S lies on the radical axis of S₁ and S₂. [Hints: Take the line of centres of S₁ and S₂ as x – axis and the radical axis as y – axis. Use conditions for the orthogonal intersection of S, S₁ and S, S₂ simultaneously and prove that S is centred on the y – axis.]
Solution:
Let the equation of the circle S, S₁ and S₂ are
x² + y² + 2gx + 2fy + C = 0      …(1)
x² + y² + 2g₁x + 2f₁y + C = 0      …(2)
and x² + y² + 2g₂x + 2f₂y + C = 0      …(3)
According to the question, the circle S intersects circles S₁ and S₂ orthogonally.
Hence 2 (g₁g + f₁f) – C₁ – C = 0 …(4)
and 2 (g₂g + f₂f) – C₂ – C = 0  ….(5)
Subtracting (4) from (3) we get
2g(g₁ – g₂) + 2f (f₁ – f₂) – C₁ + C₂ = 0 …(6)
Now radical axis of circles S₁ and S₂ is S₁ – S₂ = 0
i, e. 2x (g₁ – g₂) + 2y (f1 – f2)+ C₁ – C₂ = 0 ….(7)
The centre of the circle S is (-g, -f).
If it lies in the radical axis then equation (7) will be satisfied by the centre.
i,.e, 2g (g₁ – g₂) + 2f (f₁ – f₂) – C₁ + C₂ = 0
which is nothing but equation (5). Hence centres of S lie on the radical axis of S₁ and S₂.

Question 23.
R is the radical centre of circles S₁, S₂ and S₃. Prove that if R is on/inside/outside one of the circles then it is similarly situated with respect to the other two.
Solution:
Given R is the radical centre of S₁, S₂ and S₃
The radical centre is the intersection point of three radical axes whose equations are
S₁ – S₂ = 0
S₂ – S₃ = 0    …..(1)
S₃ – S₁ = 0
Let S₁ : x² +y² + 2g₁x + 2f₁y + C₁ =0
S₂ : x² + y² + 2g₂x + 2f₂y + C₂ =0
S₃ : x² + y² + 2g₃x + 2f₃y + C₃ =0
Now equations of radical axes by set of equation (1) are
2x(g₁ – g₂) + 2y(f₁ – f₂) + C₁ – C₂ =0 …(2)
2x(g₂ – g₃) + 2y(f₂ – f₃) + C₂ – C₃ =0 …(3)
and 2x(g₃ – g₁) + 2y(f₃ – f₁) + C₃ – C₁ = 0 …(4)
Let the co-ordinate of R be (x1, y1) the
point R must satisfy (2), (3) and (4).
i.e., 2x₁(g₁ – g₂) + 2y₁(f₁ – f₂) + C₁ – C₂ = 0 …(5)
2x₁(g₂ – g₃) + 2y₁(f₂ – f₃) + C₂ – C₃ =0 …(6)
2x₁(g₃ – g₁) + 2y₁(f₃ – f₁) + C₃ – C₁ =0 …(7)
Subtracting (6) for (5) we get
2x₁(g₁ – g₃) + 2y₁(f₁ – f₃) + C₁ – C₃ =0
⇒ 2g₁x₁ + 2f₁y₁ + C₁ =2g₃x₁ + 2f₃y₁ + C₃
Similarly subtracting (7) from (6) we get
2g₂y₁ + 2f₂y₁ + C₂ = 2g₁x₁ + 2f₁y₁ + C₁
Combining the above two equations we get
2g₂x₁ + 2f₂y₁ + C₂ = 2g₁x₁ + 2f₁y₁ + C₁ = 2g₃x₁ + 2f₃y₁ + C₃
If R x₁ y₁ lies on / inside / outside of S₁ …(8) then x₁² + y₁² + 2g₁x₁ + 2f1y1 + C₂ (= / < / >)0 respectively.
⇒ x₁² + y₁² + 2g₂x₂ + 2f₂y₂ + C₂(=/</>) 0
⇒ x₁² + y₁² + 2g₃x₃ + 2f₃y₃ + C₃(=/</>) 0
respectively by Eqn (8).
This concludes that if R is on /inside/outside. One of the circles then it is similarly situated with respect to the other two.

Question 24.
Determine a circle which cuts orthogonally to each of the circles.
S₁: x² + y² + 4x – 6y + 12 = 0
S₂: x² + y² + 4x + 6y + 12 = 0
S₃: x² + y² – 4x + 6y + 12 = 0
[Hints: The centre of the required circle S must be the radical centre R (why?), which lies outside all the circles. Then show that the radius of S must be the length of the tangent from R to any circle of the system.
Solution:
Let the equation of the required circle is x² + y² + 2gx + 2fy + C = 0    …..(1)
We know if two circles
x² + y² + 2g₁x + 2f₁y + C₂ = 0 and
x² + y² + 2g₂x + 2f₂y + C₂ = 0
are orthogonal then
2(g₁g₂ + f₁f₂) – C₁ – C₂ =0   ….(2)
According to the question circle (1) is orthogonal to the circles
S₁: x² + y² – 4x – 6y + 12 = 0     ….(3)
S₂: x² + y² + 4x + 6y + 12 = 0   ….(4)
S₃: x² + y² – 4x + 6y + 12 = 0     ….(5)
For these circles equation (2) will be
2(-2g – 3f) – C – 12 = 0     ….(6)
2(2g + 3f) – C – 12 = 0     …..(7)
2(-2g + 3f) – C – 12 = 0     …..(8) respectively.
Now subtract eqn. (7) from (6) and (8) from (7) we get
2(- 4g – 6f) = 0
⇒ 2(4g) = 0 ⇒ g = 0 , and f = 0
Using the value of g and f in eq. (6) we get
C = -12
Using g = 0, f= 0 , C = -12 in (1) we get
x² + y² – 12 = 0 is the required equation of the circle.

Question 25.
Prove that no pair of concentric circle can have radical axes.
Solution:
Let the centre of pair of concentric circles is C (h, k) and radii are r₁ and r₂.
So equation of the circles are
S₁: (x – h)² + (y – k)² = r₁²
S₂: (x – h)² + (y – k)² = r₂²
Equation of the radical axis is S₁ – S₂ = 0
⇒ r₁² – r₂² = 0
which is not a straight line as r₁ and r₂ are constants.
Hence it concludes that no pair of concentric circles have a radical axis.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 6 The Past Simple and the Past Perfect Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 6 The Past Simple and the Past Perfect

SECTION – 1

Look at the sentences below.
(a) I reached the hostel in the morning and found that somebody had broken into my room during the night.
(b) She said that her friend had published a book.
(c) He had lived in this town for ten years; then he migrated to Japan.
Can you find the Past Perfect Tense in each sentence? Note that the sentence in which it occurs refers to two actions — the action expressed by the Past Perfect and another action expressed by the Past Simple. Of the two actions, which takes place earlier and which takes place later? List them below.

(a) (1) I reached the hotel in the morning and found. (later).
(2) that somebody had broken (earlier action) into my room during the night.
(b) (1) She said _________(later action).
(2) that her friend had published a book _________(earlier action).
(c) (1) _________then he migrated to Japan _________(later action).
(2) He had lived in this town for ten years, (earlier action).
Can you answer now? Which action does the Past Perfect refer to — the earlier one or the later one? Which action does the Past Simple refer to?
Answer:
The earlier action refers to Past Perfect and the later one refers to Past Simple Tense.

Activity – 23
Combine each pair of sentences below into a single sentence, using the Past Perfect to show which action took place earlier. (You may have to use words like after; when etc.
(a) (i) I finished my homework.
(ii) Then I went to buy a pen.
_____________________________________.
(b) (i) Then the doctor gave some medicine to the patient,
(ii) Then the patient regained his senses.
_____________________________________.
(c) (i) I read a few pages from the book.
(ii) After that I returned it to the librarian.
_____________________________________.
(d) (i) I worked in the garden for some time.
(ii) After that I had my breakfast.
____________________________
(e) (i) He left the place in a hurry.
(ii) After that his friend arrived.
____________________________
(f) (i) The young girl finished shopping.
(ii) Then she met with an accident.
____________________________
(g) (i) The thief ran away with the gold.
(ii) After that the police arrived.
____________________________
Answer:
(a) After I had finished my homework, I went to buy a pen.
(b) After the doctor had given some medicine to the patient, the patient regained his senses.
(c) After I had read a few pages from the book, I returned to the librarian.
(d) I had worked in the garden for some time before I had my breakfast.
(e) When he had left the place in a hurry. his friend arrived.
(f) After the young girl had finished shopping. she met with an accident.
(g) The thief had run away with the gold before the police arrived.

Activity — 24
Jatin arrived late at different places yesterday. What did he find when he arrived
at each place?
Example — When he arrived at the cricket stadium, the game had ended.
(a) the hank it / already I close.
____________________________
(b) his uncle’s house his uncle I go the sleep.
____________________________
(c) the bus stops the bus I already / leave.
____________________________
(d) book shop the book he wanted/sold out already.
_________________________________
(e) the club his friends/leave.
_________________________________
(f) the hostel everyone / go to bed.
_________________________________
Answer:
(a) When he arrived at the bank, it had already closed.
(b) When he got to his uncle’s house, his uncle had gone to sleep.
(c) When he reached the bus stop, the bus had already left.
(d) When he came to the bookshop, the book he wanted had been sold out already.
(e) When he arrived at the club, his friends had left.
(f) When he came to the hostel, everyone had gone to bed.

Activity – 25
Use the verb supplied in brackets in the appropriate form.
(a) We went to Anil’s house and _____________(knock) on the door but there _____________ (be) no answer. Either he _____________ (go) out or he _____________ (not want) to see anyone.
(b) Sadhan _____________(go) for a walk yesterday because the doctor _____________(tell) him last week that he _____________(need) exercise.
(c) A : _____________(Seema / arrive) at the party in time last night ?
B: No, she was late. By the time, we got there, everyone _____________(leave).
Answer:
(a) We went to Anil’s house and knocked on the door but there was no answer. Either he had gone out or he did not want to see anyone.
(b) Sadhan went for a walk yesterday because the doctor told him last week that he needed exercise.
(c) A: Did Seema arrive at the party in time last night?
B: No, she was late. By the time, we got there, everyone had left.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 5 Past Simple and Past Progressive Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 5 Past Simple and Past Progressive

SECTION – 1

Study the sentences below :
(a) It started to rain while we were walking home.
(b) My sister was tidying my room when I saw your letter.
(c) Anita was walking along the road when suddenly she heard footsteps behind her. Someone was following her. She was frightened and started to run.
What do you think the use of the Past Simple and the Past Progressive indicates in these sentences?
(Hint: Think of a point of time and a duration of time in the past and relate them to the action.)

Note:
The Past Simple tells us that the work/action started and finished in the past. The speaker has a definite time in mind. But in Past Progressive the time of beginning or completion of the activity is not mentioned. The activity was in progress for that hour.

Activity – 21
Put the verbs into the correct form, Past Progressive or Past Simple.
(a) My friend ______________(meet) Anima and Amiya at the bus stop four days ago. They ______________(go) to Paradeep and my friend ______________ (go) to Bolangir. They ______________(have) a chat while they ______________(wait) for their buses.
(b) My brother ______________ (cycle) to school last Monday when suddenly an old woman ______________(step) out into the road in front of him. He ______________(go) quite fast but luckily he ______________ (manage) to stop in time and ______________(not / hit) her.
Answer:
(a) My friend met Anima and Amiya at the bus stop four days ago. They were going to Paradeep and my friend was going to Bolangir. They had a chat while they were waiting for their buses.
(b) My brother was cycling to school last Monday when suddenly an old woman stepped out into the road in front of him. He was going quite fast but luckily he managed to stop in time and did not hit her.

Activity – 22
Here is a true story.
An old couple …………………………living in a flat in Bhubaneswar. ………………………… locked up in one room ………………………….. Some unknown people took away everything ………………………… police arrived ………………………… climbed ………………………… rescued ………………………… broke open a door ………………………… one dacoit was killed ………………………… detective was called …………………………interviewed a witness.
Imagine that you are being questioned by the police as if you were a witness to the crime. A police Inspector is recording your statements in a notebook. Think about the situation and write the appropriate answers.
Inspector: Where were you standing at that time?
Answer: _____________________________.
Inspector: Why did you come here?
Answer: _____________________________.
Inspector: What was the old man doing at the time?
Answer: _____________________________.
Inspector: How did you see that?
Answer: _____________________________.
Inspector: How long were you standing there?
Answer: _____________________________.

Answer:
An old couple _____ living in a flat in Bhubaneswar, _____ Orissa locked up in one room _____. Some unknown people took away everything. _____ police arrived _____, limbed _____ rescued _____ , broke open a door _____one dacoit killed, _____detective was called _____ interviewed a witness.
Inspector: Where were you standing at that time?
Answer: I was standing near the flat.
Inspector: Why did you come here?
Answer: I came here to play.
Inspector: What was the old man doing at that time?
Answer: The old man was shouting and trembling out of fear.
Inspector: How did you see that?
Answer: I heard him shouting and saw him trembling.
Inspector: How long were you standing there?
Answer: I stood there till your arrival.

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Tense Patterns Unit 4 Present Perfect and Past Simple Textbook Activity Questions and Answers.

CHSE Odisha 12th Class English Grammar Tense Patterns Unit 4 Present Perfect and Past Simple

Study the dialogue given below.
Susant: Have you ever ridden a horse?
Subir: Yes, I have.
Susant: When was that?
Subir: I rode one last summer.
Susant : What was it like?
Subir: Oh, it was awful.
Susant : Why? What happened?
Subir: I fell off and hurt my back.
Identify the Present Perfect and Past Simple sentences and examine their use carefully. How are they different in meaning?
{Hint: One of them answers the question ‘When’? and the other does not)
Except for the first two i.e. “Have you ever ridden a horse ?” and “Yes, I have”, the rest of the sentences in the above dialogue belong to Past Simple constructions.
When an action/event took place in the past but its result is still operative at the present moment of speaking/time, we generally use a Present Perfect tense and the Past Simple means that the action/happening occurred before the present moment.

Activity – 18
1. Complete the dialogue using the hints given.
(i) A: ever / see /a lion _______________?
B: Yes, _______________.
A: Where _______________?
B: In the zoo _______________.
A: What/look _______________?
B: terrible _______________.
A: You / afraid _______________?
B: No, _______________?

(ii) A : ever / be to / Dhauligiri ………………………?
B: Yes, _______________.
A: What! see /there _______________?
B: A temple I top/hill _______________.
A See / the inscriptions _______________?
B: Yes, _______________?
A: Able to read the inscriptions _______________?
B: No, _______________.

Answer:
(i) A: Have you ever seen a lion?
B: Yes, I have.
A: Where did you see it?
B: I saw it in the zoo.
A: What did it look like?
B: Yes, it was very terrible to look at.
A: Were you afraid?
B: No, I wasn’t.

(ii) A: Have you ever been to Dhauligiri?
B: Yes, I have been two times.
A: What did you see there?
B: I saw a temple at the top of the hill.
A: Did you see the inscriptions there?
B: Yes, I saw the inscriptions there.
A: Were you able to read the inscriptions there?
B: No, I wasn’t.

Activity – 19
Choose the right verb for each blank space and put it into the correct tense.
(do, wear, carry, ask, say, think)
A : _______________your grandfather _______________ something really crazy ?
B: He _______________ something really silly last summer. One one of the hottest days he _______________ a raincoat and _______________ an umbrella. Everyone _______________ him why. He _______________he _______________ it was going to rain.
Answer:
A: Did your grandfather wear something really crazy?
B: He wore something really silly last summer. On one of the hottest days, he wore a raincoat and carried an umbrella. Everyone asked him why. He said he thought it was going to rain.

Activity – 20
Complete the sentences, using the verbs in brackets either in Past Simple or Present Perfect form.
(a) She _______________ up her mind (made). She’s going to look for another college.
(b) Amulya : _______________me his pen but I’m afraid I _______________ it. (give, lose)
(c) A: It’s a little bit noisy in here, isn’t it?
B: Pardon? I can’t hear. What _______________ you _______________? (say)
(d) Where is my bike? It _______________ outside the classroom. It _______________! (be, disappear)
(e) Did you know that Umesh _______________ a new scooter? (buy)
(f) I did Sanskrit at school but I _______________ most of it. (forget)
(g) A : Sima, this is Rajesh.
B: Hello, Rajesh. Actually, we know each other. We _______________ already ___________ (meet).

Answer:
(a) She has made up her mind. She’s going to look for another college.
(b) Amulya gave me his pen but I’m afraid I have lost it.
(c) A: It’s a little bit noisy in here, isn’t it?
B: Pardon? I can’t hear. What did you say?
(d) Where is my bike? It was outside the classroom. It has disappeared!
(e) Did you know that Umesh has bought a new scooter?
(f) I did Sanskrit at school but I have forgotten most of it.
(g) A : Sima, this is Rajesh.
B: Hello, Rajesh. Actually, we know each other. We have already met.