Question

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 ​

Answer 1

$\huge\bf{Answer:-}$

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 :

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

Answer 2

$\huge\boxed\boxed{Answer}$

${1 - 6}^{2} + {3 + 1}^{2} = {x - 1}^{2} + {8 - 3}^{2}$

$21 + 4 = {x}^{2} + 1 - 2 + 25$

${x}^{2} - 2x + 1 - 2x + 25$

${x}^{2} + x - 3x - 3 = 0$

$x \: \: x + 1 - 3 \: \: x + 1 = 0$

$x + 1 = 0 \: \: \: x - 3 = 0$

$x = - 1 \: \: \: or \: \: \: = 3$

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