# 10th CBSE math solution for exercise 7.1 part 1

This page 10th CBSE math solution for exercise 7.1 part 1 is going to provide you solution for every problems that you find in the exercise no 7.1

## 10th CBSE math solution for exercise 7.1 part 1

(1) Find the distance between the following pairs of points:

(i) (2 , 3) (4 , 1)

Distance between two points = √(x₂ - x₁) ² + (y₂ - y₁) ²

Here x₁ = 2, y₁ = 3, x₂ = 4  and  y₂ = 1

= √(4 - 2)² + (1 - 3)²

= √2² + (-2)²

= √(4 + 4)

= √8

= √(2 x 2 x 2)

= 2 √2

(ii) (-5 , 7) (-1 , 3)

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 5, y₁ = 7, x₂ = -1  and  y₂ = 3

= √(-1 -(-5))² + (3 - 7)²

= √(-1 + 5)² + (-4)²

= √4² + (-4)²

= √16 + 16

= √32

= √(2 x 2 x 2 x 2 x 2)

= 2 x 2 √2

= 4 √2

(iii) (a , b) (-a , -b)

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = a, y₁ = b, x₂ = -a  and  y₂ = -b

= √(-a -a)² + (-b - b)²

= √(-2a)² + (-2b)²

= √4a² + 4b²

= √4(a² + b²)

= √2 x 2(a² + b²)

= 2 (a² + b²)

(2) Find the distance between the points (0,0) and (36,15). Can you find the distance between the two points A and B discussed in section 7.2

Solution:

Let A (0,0) B (36,15)

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 0, y₁ = 0, x₂ = 36  and  y₂ = 15

= √(36 - 0)² + (15 - 0)²

= √(36)² + (15)²

= √1296 + 225

= √1521

= √39 x 39

= 39

So the distance between the given towns A and B will be 39 km.

In the page 10th CBSE math solution for exercise 7.1 part 1 we are going to see the solution of next problem

(3) Determine if the points (1,5) (2,3) and (-2,-11) are collinear.

Solution:

A (1,5)  B(2,3) and C (-2,-11)

Distance between A and B

Here x₁ = 1, y₁ = 5, x₂ = 2  and  y₂ = 3

AB =  √(2 - 1)² + (3 - 5)²

=  √(1)² + (- 2)²

=  √1 + 4

=  √5

Distance between B and C

Here x₁ = 2, y₁ = 3, x₂ = -2  and  y₂ = -11

AB =  √(-2 - 2)² + (-11 - 3)²

=  √(-4)² + (-14)²

=  √16 + 196

=  √212

Distance between C and A

Here x₁ = -2, y₁ = -11, x₂ = 1  and  y₂ = 5

AB =  √(1 -(-2))² + (5 -(-11))²

=  √(1+2)² + (5 + 11)²

=  √3² + 16²

=  √9 + 256

=  √265

Since AB + BC CA