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

Step-by-step explanation:

P = (2, 5)

Q = (x, -3)

R = (7, 9)

$d_{PQ}=d_{QR}$

$d_{PQ}=\sqrt{(2-x)^2+(5+3)^2}, \quad d_{QR} = \sqrt{(x-7)^2+(-3-9)^2}$

$\sqrt{(2-x)^2+(5+3)^2} = \sqrt{(x-7)^2+(-3-9)^2}$

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

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

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

$10x+68 = 193$

$10x= 125$

x = 12.5