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

(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 + y****² + 6 y + 9 = 100**

**y****² + 6 y + 9 + 64 - 100 = 0**

**y****² + 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 + x****² - 4 x + 25 = 4 + x****² + 4 x + 81**

** ****x****² - ****x****² - 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****²**-6x+9+y** ²-12y+36) **=

**taking squares on both sides **

**x****²-****x****²+y****²-y****²-6x-6x-12y+8y+45-25=0**

**-12x-4y+20 = 0**

**divide the whole equation by (-4)**

** 3 x + y - 5 = 0**

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