Question

Find the value of x such that PQ = QRwhere P, Q and Rare the pints (2, 5), (x, -3) and (7, 9) respectively.

Answer 1

the value of x is x,-3

Answer 2

Answer: x = 12.5

Explanation:

P = (2, 5)

Q = (x, -3)

R = (7, 9)

`d_(PQ)=d_(QR)`

`d_(PQ)=√((2-x)^2+(5+3)^2), \quad d_(QR) = √((x-7)^2+(-3-9)^2)`

`√((2-x)^2+(5+3)^2) = √((x-7)^2+(-3-9)^2)`

`(2-x)^2+(5+3)^2 = (x-7)^2+(-3-9)^2`

`4-4x+x^2+64 = x^2-14x+49+144`

`4x+68 = -14x+193`

`10x+68 = 193`

`10x= 125`

x = 12.5