10th mathematics exercise questions with answers

# Coordinate Geometry Exercise 14.2 – Class 10

### Coordinate Geometry Exercise 14.2 – Questions:

1. Find the equation of the line whose angle of inclination and y-intercept are given.

(i) θ = 60˚, y-intercept is -2.

(ii) θ = 45˚, y-intercept is 3.

1. Find the equation of the line whose slope and y-intercept are given.

(i) Slope = 2, y-intercept = -4

(ii) Slope = –2/3, y-intercept = 1/2

(iii) Slope = -2, y-intercept = 3

1. Find the slope and y-intercept of the lines

(i) 2x + 3y = 4

(ii) 3x = y

(iii) x – y + 5 = 0

(iv) 3x – 4y = 5

1. Is the line x = 2y parallel to 2x – 4y + 7 = 0[Hint: Parallel lines have some slopes]
2. Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each other.[Hint: For perpendicular lines, m1.m2 = -1].

## Coordinate Geometry Exercise 14.2 – Solutions:

1. Find the equation of the line whose angle of inclination and y-intercept are given.

(i) θ = 60˚, y-intercept is -2.

Solution:

m = tan θ = tan 60˚ = √3

y – intercept = c = -2

Equation: y = mx + c

y = √3x – 2

(ii) θ = 45˚, y-intercept is 3.

Solution:

m = tan θ = tan 45˚ = 1

y – intercept = c = 3

Equation: y = mx + c

y = x + 3

1. Find the equation of the line whose slope and y-intercept are given.

(i) Slope = 2, y-intercept = -4

Solution:

m = slope = 2

y – intercept = c = -4

Equation: y = mx + c

y = 2x – 4

(ii) Slope = –2/3, y-intercept = 1/2

Solution:

m = slope = –2/3

y – intercept = c = 1/2

Equation: y = mx + c

y = –2/3 x + 1/2

(iii) Slope = -2, y-intercept = 3

Solution:

m = slope = -2

y – intercept = c = 3

Equation: y = mx + c

y = -2x + 3

1. Find the slope and y-intercept of the lines

(i) 2x + 3y = 4

Solution:

2x + 3y = 4

3y =  4 – 2x

y = – 2/3x + 4/3

Standard equation is y = mx + c

Therefore, m = –2/3 and c = 4/3

(ii) 3x = y

Solution:

Given, 3x = y

y = 3x + 0

Standard equation is y = mx + c

Therefore, m = 3 and c = 0

(iii) x – y + 5 = 0

Solution:

x – y + 5 = 0

-y = – x – 5

y = x + 5

Standard equation is y = mx + c

Therefore, m = 1 and c = 5

(iv) 3x – 4y = 5

Solution:

3x – 4y = 5

-4y = 5 – 3x

4y = 3x – 5

y = 3/4 x – 5/4

Standard equation is y = mx + c

Therefore, m = 3/4 and c = – 5/4

1. Is the line x = 2y parallel to 2x – 4y + 7 = 0[Hint: Parallel lines have same slopes]

Solution:

line 1:

x = 2y

y = 1/2 x + 0

Standard equation is y = mx + c

Therefore, m1 = 1/2 and c1 = 0

line 2:

2x – 4y + 7 = 0

-4y = 7 – 2x

4y = 2x – 7

y = 2/4 x – 7/4

Standard equation is y = mx + c

Therefore, m2 = 2/4 = 1/2 and c2 = –7/4

Since slope of the line x = 2y and the line 2x – 4y + 7 = 0 are same i.e., m1 = m2 = 1/2.

Therefore, the line x = 2y parallel to 2x – 4y + 7 = 0.

1. Show that the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular to each other.[Hint: For perpendicular lines, m1.m2 = -1].

Solution:

line 1:

3x + 4y + 7 = 0

4y = – 3x – 7

y = – 3/4 x – 7/4

Standard equation is y = mx + c

Therefore, m1 = – 3/4 and c1 = –7/4

line 2:

28x – 21y + 50 = 0

-21y = – 28x – 50

21y = 50 + 28x

y = 28/21 x + 50/21

Standard equation is y = mx + c

Therefore, m2 = 28/21 and c2 = 50/21

If slope of the line 3x + 4y + 7 = 0 and 28x – 21y + 50 = 0 are perpendicular then we must get m1.m2 = -1

i.e.,

3/4 . 28/21 = -1

Therefore, the line 3x + 4y + 7 = 0 is perpendicular to 28x – 21y + 50 = 0.