Find the value of x pQ=QR the coordinates of p,q and r and (6,-1),(1,3)and(x,8) respectively.

Answer 1

If PQ = QR , (where P(6,-1), Q(1,3), R(x,8))

√( (3+1)^2 + (1-6)^2) = √( (8-3)^2 + (x-1)^2 )
√41 = √(x^2 - 2x + 26)
square both sides
41 = x^2 - 2x + 26
x^2 - 2x - 15 = 0

(x-5)(x+3) = 0
x = 5 or x = -3

Answer 2

what a exact answer no doubt