Frank Solutions Class 9 Maths Chapter 28 Coordinate Geometry

Frank Solutions for Class 9 Maths Chapter 28 Coordinate Geometry are provided here in PDF. Coordinate Geometry is one of the most interesting concepts of Mathematics. The solutions are curated by the subject experts at BYJU’S in a lucid manner, helping students in solving geometric problems effortlessly. Accurate answers to exercise problems in the textbook build students’ interest in Mathematics and help them prepare well for their exams. For in-depth knowledge of the concepts, practise these solutions regularly.

Chapter 28, Coordinate Geometry, deals with the study of geometry using coordinate points. The solutions provide information about plotting graphs on a plane. Regular practice of these solutions will enhance the skills which are essential to score high in exams. To perform well in the annual exam, students are suggested to download Frank Solutions for Class 9 Maths Chapter 28 Coordinate Geometry from the link below and practise offline as well.

Frank Solutions for Class 9 Maths Chapter 28 Coordinate Geometry Download PDF

Access Frank Solutions for Class 9 Maths Chapter 28 Coordinate Geometry

1. Find the value of ‘a’ and ‘b’ if

(a) (a + 2, 5 + b) = (1, 6)

(b) (2a + b, a – 2b) = (7, 6)

Solution:

(a) Given,

Two ordered pairs are equal

a + 2 = 1 and 5 + b = 6

a = 1 – 2 b = 6 – 5

a = – 1 b = 1

Therefore,

a = – 1 and b = 1

(b) Given,

Two ordered pairs are equal

2a + b = 7 ……. (1)

a – 2b = 6 ……. (2)

Multiplying equation (1) with (2), we get,

4a + 2b = 14 …… (3)

On adding equation (2) and (3), we get,

a – 2b = 6

(+) 4a + 2b = 14

_________________

5a = 20

Hence,

a = 4

Substituting a = 4 in equation (1), we get,

2(4) + b = 7

8 + b = 7

b = 7 – 8

b = – 1

Therefore,

a = 4 and b = -1

2. State the quadrant in which each of the following points lie:

A (-4, 3), B (2, -5), C (-5, -3), M (4, 8), P (-1, 9) and Z (4, -5)

Solution:

The quadrant in which the following point lies are as follows:

A (-4, 3): II quadrant

B (2, -5): IV quadrant

C (-5, -3): III quadrant

M (4, 8): I quadrant

P (-1, 9): II quadrant

Z (4, -5): IV quadrant

3. State the axis on which the following points lie:

J (0, -7), M (5, 0), P (-4, 0), R (0, 6) and W (2, 0)

Solution:

The axis on which the following point lies are as follows:

J (0, -7): y-axis

M (5, 0): x-axis

P (-4, 0): x-axis

R (0, 6): y-axis

W (2, 0): x-axis

4. Find the co-ordinates of points whose:

(i) Abscissa is 6 and ordinate is 2

(ii) Abscissa is 0 and ordinate is -3

(iii) Abscissa is 5 and ordinate is -1

(iv) Abscissa is -2 and ordinate is 0

(v) Abscissa is -4 and ordinate is -7

(vi) Abscissa is 0 and ordinate is 0

(vii) Abscissa is -7 and ordinate is 4

Solution:

(i) The co-ordinates of a point whose abscissa is 6 and ordinate is 2 is (6, 2)

(ii) The co-ordinates of a point whose abscissa is 0 and ordinate is -3 is (0, -3)

(iii) The co-ordinates of a point whose abscissa is 5 and ordinate is -1 is (5, -1)

(iv) The co-ordinates of a point whose abscissa is -2 and ordinate is 0 is (-2, 0)

(v) The co-ordinates of a point whose abscissa is -4 and ordinate is -7 is (-4, -7)

(vi) The co-ordinates of a point whose abscissa is 0 and ordinate is 0 is (0, 0)

(vii) The co-ordinates of a point whose abscissa is -7 and ordinate is 4 is (-7, 4)

5. Plot the following points on the graph paper:

P (2, 5), Q (4, 0), R (0, 7), S (-3, 5), T (4, -4), U (0, -2) and V (-1, -4)

Solution:

6. Plot the points O (0, 0), P (6, 0) and R (0, 5) on a graph. Find the coordinates of point Q such that OPQR is a rectangle

Solution:

The coordinates of the point Q is (6, 5)

Hence,

Q (6, 5)

7. Plot points A (3, 4) and C (-3, -2) on a graph. Find the coordinates of points B and D such that ABCD is a square. Also, find the area of the square.

Solution:

From the figure,

B = (-3, 4) and D = (3, -2)

We know that,

Area of the square = side²

= 6²

= 36 sq. units

Therefore,

The area of the square is 36 sq. units

8. Plot the points (-2, 3), (3, 3), (5, -2) and (-5, -2) on a graph and join them in order. Name the figure you get.

Solution:

The quadrilateral obtained by joining the given points on a graph is a trapezium

9. Express the equation 3x + 5y + 15 = 0 in the form such that:

(a) x is subject to the formula

(b) y is dependent variable and x is independent variable

Solution:

(a) 3x + 5y + 15 = 0

On calculating further, we get,

3x = -5y – 15

x = (-5y – 15) / 3

We get,

x = (-5 / 3) y – 5

(b) 3x + 5y + 15 = 0

On further calculation, we get,

5y = – 3x – 15

y = (-3x – 15) / 5

We get,

y = (- 3/ 5) x – 3

10. Draw a graph of each of the following equations:

(a) x + 6y = 15

(b) 3x – 2y = 6

(c) 3y + 2x = 11

(d) 5x + 2y =16

(e) x + y – 3 = 0

(f) x = -3y

(g) y = (5 / 2)x + (2 / 5)

(h) {(x – 2) / 3} – {(y + 1) / 2} = 0

(i) 2 (x – 5) = (3 / 4) (y – 1)

(j) y = (3 / 5) x – 1

Solution:

(a) x + 6y = 15

On simplification, we get,

x = 15 – 6y

When y = 1,

x = 15 – 6(1)

x = 9

When y = 2,

x = 15 – 6(2)

x = 3

When y = 3,

x = 15 – 6 (3)

x = -3

Plotting the points (9, 1), (3, 2) and (-3, 3), we get the line segment as shown in the figure below

(b) 3x – 2y = 6

On calculating further, we get,

2y = 3x – 6

y = (3x – 6) / 2

When x = 2,

y = {3(2) – 6} / 2

y = 0

When x = 4,

y = {3 (4) – 6} / 2

y = (12 – 6) / 2

y = 6 / 2

y = 3

When x = -2,

y = {3(-2) – 6} / 2

y = (-12 / 2)

y = -6

Plotting the points (2, 0), (4, 3) and (-2, -6), we get a line segment as shown in the figure below

(c) 3y + 2x = 11

3y = 11 – 2x

y = (11 – 2x) / 3

When x = 1,

y = {11 – 2(1)} / 3

y = 9 / 3

y = 3

When x = -2,

y = {11 – 2(-2)} / 3

y = 15 / 3

y = 5

When x = -5,

y = {11 – 2(-5)} / 3

y = 21 / 3

y = 7

Plotting the points (1, 3), (-2, 5) and (-5, 7), we get a line segment as shown in the figure below

(d) 5x + 2y = 16

2y = 16 – 5x

y = (16 – 5x) / 2

When x = 2,

y = {16 – 5 (2)} / 2

y = 3

When x = 4,

y = {16 – 5(4)} / 2

y = -2

When x = 6,

y = {16 – 5(6)} / 2

y = -7

Plotting the points (2, 3), (4, -2) and (6, -7), we get a line segment as shown in the figure below

(e) x + y – 3 = 0

y = 3 – x

When x = 2,

y = 3 – 2

y = 1

When x = 0,

y = 3 – 0

y = 3

When x = 6,

y = 3 – 6

y = -3

Plotting the points (2, 1), (0, 3) and (6, -3), we get a line segment as shown in the figure below

(f) x = -3y

When y = 1,

x = -3 (1)

x = -3

When y = 0,

x = -3 (0)

x = 0

When y = -2,

x = -3(-2)

x = 6

Plotting the points (-3, 1), (0, 0) and (6, -2), we get a line segment as shown in the figure below

(g) y = (5 / 2) x + (2 / 5)

When x = 1,

y = (5 / 2) (1) + (2 / 5)

y = 2.9

When x = 0,

y = (5 / 2) (0) + (2 / 5)

y = 0.4

When x = 2,

y = (5 / 2) (2) + (2 / 5)

y = 5.4

Plotting the points (1, 2.9), (0, 0.4) and (2, 5.4), we get a line segment as shown in the figure below

(h) {(x – 2) / 3} – {(y + 1) / 2} = 0

{(x – 2) / 3} = {(y + 1) / 2}

2 (x – 2) = 3 (y + 1)

2x – 4 = 3y + 3

3y = 2x – 7

We get,

y = (2x – 7) / 3

When x = 2,

y = {2(2) – 7} / 3

y = (-3 / 3)

y = -1

When x = -1,

y = {2(-1) – 7} / 3

y = (-9 / 3)

y = -3

When x = -2.5,

y = {2(-2.5) – 7} / 3

y = (-12 / 3)

y = -4

Plotting the points (2, -1), (-1, -3) and (-2.5, -4), we get a line segment as shown in the figure below

(i) 2 (x – 5) = (3 / 4) (y – 1)

8 (x – 5) = 3 (y – 1)

8x – 40 = 3y – 3

3y = 8x – 40 + 3

On calculating further, we get,

3y = 8x – 37

y = (8x – 37) / 3

When x = 2,

y = {8 (2) – 37} / 3

y = {16 – 37} / 3

y = (-21 / 3)

y = -7

When x = 5,

y = {8(5) – 37} / 3

y = (3 / 3)

y = 1

When x = -1,

y = {8(-1) – 37} / 3

y = (-45 / 3)

y = -15

Plotting the points (2, -7), (5, 1) and (-1, -15), we get a line segment as shown in the figure below

(j) y = (3 / 5) x – 1

When x = 5,

y = (3 / 5) (5) – 1

y = 3 – 1

y = 2

When x = -5,

y = (3 / 5) (-5) -1

y = -3 – 1

y = -4

When x = 10,

y = (3 / 5) (10) – 1

y = 6 – 1

y = 5

Plotting the points (5, 2), (-5, -4) and (10, 5), we get a line segment as shown in the figure below

11. Draw a graph for each of the following equations and find the coordinates of the points where the line drawn meets the x-axis and y-axis:

(a) 2x + 3y = 12

(b) (2x / 5) + (y / 2) = 1

Solution:

(a) 2x + 3y = 12

3y = 12 – 2x

y = 4 – (2 / 3) x

When x = 3,

y = 4 – (2 / 3) (3)

y = 4 – 2

y = 2

When x = -3,

y = 4 – (2 / 3) (-3)

y = 4 + 2

y = 6

When x = 6,

y = 4 – (2 / 3) (6)

y = 4 – 4

y = 0

Plotting the points (3, 2), (-3, 6) and (6, 0), we get a line segment as shown in the figure below

The coordinates of the points where the line meets the x-axis is (6, 0), and the y-axis is (0, 4)

(b) (2x / 5) + (y / 2) = 1

(y / 2) = 1 – (2x / 5)

(y / 2) = (5 – 2x) / 5

We get,

y = (10 – 4x) / 5

When x = 0,

y = {10 – 4 (0)} / 5

y = 10 / 5

y = 2

When x = 5,

y = {10 – 4(5)} / 5

y = (10 – 20) / 5

y = (-10 / 5)

y = -2

When x = (5 / 2),

y = {10 – 4(5 / 2)} / 5

y = 0

Plotting the points (0, 2), (5, -2) and (5/2, 0), we get a line segment as shown in the figure below

The coordinates of the points where the line meets the x-axis is (5/ 2, 0), and the y-axis is (0, 2)

12. Draw the graph of the lines y = x + 2, y = 2x – 1 and y = 2 from x = -3 to 4, on the same graph paper. Check whether the lines drawn are parallel to each other.

Solution:

For,

y = x + 2

When x = 0,

y = 0 + 2

y = 2

When x = 5,

y = 5 + 2

y = 7

When x = -3,

y = -3 + 2

y = -1

For,

y = 2x – 1

When x = 0,

y = 2(0) – 1

y = -1

When x = -2,

y = 2(-2) -1

y = -4 -1

y = -5

When x = 3,

y = 2(3) – 1

y = 6 – 1

y = 5

For,

y = 2

This line is parallel to the x-axis and passes through (0, 2)

The lines are not parallel to each other

13. Find the slope of the line whose inclination is given as

(a) 0⁰

(b) 30⁰

(c) 45⁰

(d) 60⁰

Solution:

(a) Slope = tan θ

= tan 0⁰

We get,

= 0

Hence, the slope of the line is 0

(b) Slope = tan θ

= tan 30⁰

We get,

= (1 / √3)

Hence, the slope of the line is (1 / √3)

(c) Slope = tan θ

= tan 45⁰

We get,

= 1

Hence, the slope of the line is 1

(d) Slope = tan θ

= tan 60⁰

We get,

= √3

Hence, the slope of the line is √3

14. Find the inclination of the line whose slope is:

(a) 1

(b) √3

Solution:

(a) Slope = tan θ

1 = tan θ (given)

We know that,

tan 45⁰ = 1

tan θ = tan 45⁰

Therefore,

θ = 45⁰

The inclination of the line is 45⁰

(b) Slope = tan θ

√3 = tan θ (given)

We know that,

tan 60⁰ = √3

tan θ = tan 60⁰

Therefore,

θ = 60⁰

The inclination of the line is 60⁰

15. Find the slope and y-intercept for each of the following equations:

(a) 3x – 8y + 24 = 0

(b) 6x = 7y – 12

Solution:

(a) 3x – 8y + 24 = 0

8y = 3x + 24

y = (3 / 8) x + (24 / 8)

We get,

y = (3 / 8) x + 3

Hence,

Slope = (3 / 8) and intercept = 3

(b) 6x = 7y – 12

7y = 6x + 12

We get,

y = (6 / 7) x + (12 / 7)

Hence,

Slope = (6 / 7) and intercept = (12 / 7)