# 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**

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**

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**

**Answer**

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 = side^{2}

= 6^{2}

= 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**

**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

**(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

**(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

**(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

**(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

**(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

**(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

**(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

(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

**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

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)

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

**(a) 0**°

**(b) 30**°

**(c) 45**°

**(d) 60**°

**Answer**

**(a)** Slope = tan Î¸

= tan 0°

We get,

= 0

Hence, slope of the line is 0

**(b)** Slope = tan Î¸

= tan 30°

We get,

= 1/√3

Hence, slope of the line is 1/√3

**(c)** Slope = tan Î¸

= tan 45°

We get,

= 1

Hence, slope of the line is 1

**(d) **Slope = tan Î¸

= 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 Î¸

1 = tan Î¸ **(given)**

We know that,

tan 45° = 1

⇒ tan Î¸ = tan 45°

Therefore,

Î¸ = 45°

The inclination of the line is 45^{0}

**(b)** Slope = tan Î¸

√3 = tan Î¸ **(given)**

We know that,

tan 60° = √3

⇒ tan Î¸ = tan 60°

Therefore,

Î¸ = 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)

**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) x, when y = 3 **

**(ii) y, when x = 0 **

**Answer**

**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)

**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**

**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)

**(c)**

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

**(a) 0****°**

**(b) 30°**

**(c) 45°**

**(d) 60°**

**Answer**

(a)

(b)

**(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)

**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**