Question

Find the value of x such that PQ=QR,where the coordinate of P,Q and R are (6,_1),(1,_3) and (x,8) respectively

Answer 1

mistake Q(1,3)Hope it helps! mark as brainliest

Answer 2

$\huge\bf{Answer:-}$

Given:

find the value of x such that PQ is equal to QR where the coordinates of p 6, -1 Q 13 and r x 8 respectively .

Hence :

$\sf = (1 - 6) {}^{2} + (3 + 1) {}^{2} = (x - 1) {}^{2} + (8 - 3) {}^{2} \\ \sf = 21 + 4 = x {}^{2} + 1 - 2 + 25 \\ \sf = x {}^{2} - 2x + 1 - 2x + 25 \\ \sf = x {}^{2} + x - 3x - 3 = 0 \\ \sf = x(x + 1) - 3(x + 1) = 0 \\ \sf = x + 1 = 0..x - 3 = 0 \\ \sf = x = - 1 \: \: \: \: or \: \: \: \: = 3$