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




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