Frank Solutions for Chapter 28 Coordinate Geometry Class 9 Mathematics ICSE

Exercise 28.1


1. Find the value of ‘a’ and ‘b’ if

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

(b) (2a + b, a – 2b) = (7, 6)

Answer

(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) + (4a + 2b) = 6 + 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 point lies:

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

Answer

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)

Answer

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

Answer

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

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

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

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

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

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

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

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


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)

Answer


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

Answer

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

Hence,

Q (6, 5)


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

Answer:

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.

Answer

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

Answer

(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

Answer

(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 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

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

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

Answer

(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 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 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.

Answer

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) 0°

(b) 30°

(c) 45°

(d) 60°

Answer

(a) Slope = tan θ

= tan 0°

We get,

= 0

Hence, slope of the line is 0

(b) Slope = tan θ

= tan 30°

We get,

= 1/√3

Hence, slope of the line is 1/√3

(c) Slope = tan θ

= tan 45°

We get,

= 1

Hence, slope of the line is 1

(d) Slope = tan θ

= tan 60°

We get,

= √3

Hence, slope of the line is √3


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

(a) 1

(b) √3

Answer

(a) Slope = tan θ

1 = tan θ (given)

We know that,

tan 45° = 1

⇒ tan θ = tan 45°

Therefore,

θ = 45°

The inclination of the line is 450

(b) Slope = tan θ

√3 = tan θ (given)

We know that,

tan 60° = √3

⇒ tan θ = tan 60°

Therefore,

θ = 60°

The inclination of the line is 60°


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

(a) 3x – 8y + 24 = 0

(b) 6x = 7y – 12

Answer

(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)


Exercise 28.2


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

(a) x + 5

(b) y – 4

(c) 2x = 7

(d) 2x = 7

(d) 2y – 5 = 0

(e) x = 0

(f) y = 3

Answer

(a) x + 5 = 0

⇒ x = - 5

(b) y – 4 = 0

⇒ y = 4

(c) 2x = 7

⇒ x = 7/2

(d) 2y – 5 = 0

⇒ y = 5/2

(e) x = 0

(f) y = 3


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

(a) x + 6y = 15

(b) 3x – 2y = 6

(c) 3y + 2y = 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

Answer

(a)

(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)


3. Draw a graph of the equation 3x – y = 7. From the graph find the value of :

(i) y, when = 1

(ii) x, when y = 8

Answer

(i)

(ii)


4. Draw a graph of the equation 2x – 3y = 15. From the graph find the value of:

(i) x, when y = 3

(ii) y, when x = 0

Answer

(i)

(ii)

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

(a) 2x + 3y = 12

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

Answer

(a)

(b)

The line meets the x-axis at (5/2, 0) and y-axis at (0, 2).


6. Draw a graph of the equation 2x + 3y + 5 = 0. From the graph find the value of:

(i) x, when y = - 3

(ii) y, when x = 8

Answer


7. Draw a graph of the equation 5x – 3y = 1. From the graph find the value of:

(i) x, when y = 8

(ii) y, when x = 2

Answer

(i)

(ii)


8. 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.

Answer


9. Find the inclination and slope of a line which is

(a) equidistant from the x-axis.

(b) equidistant from y-axis.

(c) intersecting x-axis at right angle.

(d) perpendicular to y-axis.

Answer

(a)

(b)
(c)
(d)

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

(a) 0°

(b) 30°

(c) 45°

(d) 60°

Answer

(a)

(b)

(c)
(d)


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

(a) 1

(b) √3

Answer

(a)

(b)


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

(a) 3x – 8y + 24 = 0

(b) 6x = 7y – 12

Answer

(a)

(b)


13. Find the equation of the line, whose

(a) slope is 3 and y-intercept 5.

(b) slope is 0 and y-intercept is – 1.

(c) slope is 1 and y-intercept is 0.

Answer

(a)

(b)
(c)


14. Draw the graph of a line 2x + 3y = 6. From the graph, find the slope and y-intercept of the line.

Answer


15. Draw the graph of the lines represented by the equations x + y = 4 and 2x – y = 2 on the same graph. Find the coordinates of the point where they intersect.

Answer


16. Draw the graph of the lines represented by the equations 3x – 2y = 4and x + y = 3 on the same graph. Find the coordinates of the point where they intersect. State, whether the lines are perpendicular to each other.

Answer


17. Draw the graph of the lines represented by equations 2x – y = 8 and 4x + 3y = 6 on the same graph. Find the co-ordinates of the point where they intersect.

Answer


18. Draw the graph of the lines represented by the equations by 5y = 3x + 1 and y = 2x + 3 on the graph. Find the coordinates of the point where they intersect.

Answer

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