# 10th CBSE math solution for exercise 7.1 part 5

This page 10th CBSE math solution for exercise 7.1 part 5 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 5

(8) Find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units.

Solution:

Distance between PQ = 10 units

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

Here x₁ = 2, y₁ = -3, x₂ = 10  and  y₂ = y

= √(10-2)² + (y-(-3))²

√(8)² + (y + 3)² = 10

64 + y² + 6 y + 9 = 100

y² + 6 y + 9 + 64 - 100 = 0

y² + 6 y - 27 = 0

(y + 9) (y - 3) = 0

y + 9 = 0        y - 3 = 0

y = -9          y = 3

In the page 10th CBSE math solution for exercise 7.1 part 5 we are going to see the solution of next problem

(9) If Q(0,1) is equidistant from P(5,-3) and R(x,6),find the values of x.Also find the distances QR and PR.

Solution:

Distance between PQ = QR

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

Here x₁ = 5, y₁ = -3, x₂ = 0  and  y₂ = 1

PQ   = √(0-5)² + (1-(-3))²

= √(2 - x)² + 25

Here x₁ = x, y₁ = 0, x₂ = -2  and  y₂ = 9

= √(-2-x)² + (9-0)²

= √(2+x)² + (9)²

= (2+x)² + 81

√(2 - x)² + 25 = (2+x)² + 81

4 + x² - 4 x + 25 = 4 + x² + 4 x + 81

x² - x² - 4 x - 4 x + 4 - 4 = 81 - 25

-8 x = 56

x = -7

Therefore the required point is (-7,0)

(10) Find a relation between x and y such that the points (x,y) is equidistant from the points (3,6) and (-3,4).

Solution:

Point (x,y) is equidistant from the points (3,6) and (-3,4)

√(x-3)² + (y-6)² = (x+3)² + (y-4)²

√(x²-6x+9+y²-12y+36) (x²+6x+9+y²-8y+16)

taking squares on both sides

x²-x²+y²-y²-6x-6x-12y+8y+45-25=0

-12x-4y+20 = 0

divide the whole equation by (-4)

3 x + y - 5 = 0