# 10th samacheer kalvi math solution for exercise 5.3 part 1

This page 10th samacheer kalvi math solution for exercise 5.3 part 1 is going to provide you solution for every problems that you find in the exercise no 5.3

## 10th samacheer kalvi solution for exercise 5.3 part 1

1. Find the angle of inclination of the straight line whose slope is
(i) 1

Solution:

slope (m) = 1

m = tan θ

tan θ = 1

for which angle of tan we get the value 1

θ = 45°

(ii) √3

Solution:

slope (m) = √3

m = tan θ

tan θ = √3

for which angle of tan we get the value √3

θ = 60°

(iii) 0

Solution:

slope (m) = 0

m = tan θ

tan θ = 0

for which angle of tan we get the value  0

θ = 0°

2. Find the slope of the straight line whose angle of inclination is
(i) 30
°

Solution:

m = tan θ

m = tan 30°

= 1/√3

(ii) 60°

Solution:

m = tan θ

m = tan 60°

= √3

(iii) 90°

Solution:

m = tan θ

m = tan 90°

=

In the page 10th samacheer kalvi math solution for exercise 5.3 part 1 we are going to see the solution of next problem

3. Find the slope of the straight line passing through the points

(i) (3 , -2) and (7 , 2)

Solution:

m = (y₂ - y₁)/(x₂ - x₁)

x₁ = 3     x₂ = 7

y₁ = -2     y₂ = 2

m = (2 - (-2))/(7- 3)

m = (2 +2)/(7- 3)

m = 4/4

= 1

(ii) (2 , -4) and origin

Solution:

m = (y₂ - y₁)/(x₂ - x₁)

x₁ = 2     x₂ = 0

y₁ = -4     y₂ = 0

m = (0 - (-4))/(0- 2)

m = (0 + 4)/(- 2)

m = 4/(-2)

= -2

(iii) (1 + √3 , 2) and (3 + √3 , 4)

Solution:

m = (y₂ - y₁)/(x₂ - x₁)

x₁ = 1 + √3     x₂ = 3 + √3

y₁ = 2           y₂ = 4

m = [(3 + √3)- (1 + √3)]/(4 - 2)

= (3 + √3- 1 - √3)/2

= 2/2

= 1