Find the value of $x$ such that $P Q = Q R$ where the coordinates of $P, Q$ and $R$ are $(6, - 1), (1, 3)$ and $(x, 8)$ respectively.

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Solution

Coordinate of point P,Q and R are P(6,-1),Q(1,3) and R(x,8)
Given, PQ=QR

So by distance formula we have,
Distance between two points = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

PQ= $\sqrt{(6 - 1)^{2} + (- 1 - 3)^{2}} = \sqrt{25 + 14} = \sqrt{41}$

$∴ P Q^{2} = 41 = Q R^{2}$

but, $Q R^{2} = (x - 1)^{2} + 25$

$41 = x^{2} + 1 - 2 x + 25$

$\Rightarrow 41 = x^{2} + 1 - 2 x + 25$
$\Rightarrow x^{2} - 2 x + 26 = 41$
$\Rightarrow x^{2} - 2 x - 15 = 0$
$\Rightarrow x^{2} - 5 x + 3 x - 15 = 0$
$\Rightarrow x (x - 5) + 3 (x - 5) = 0$
$\Rightarrow (x + 3) (x - 5) = 0$

$\Rightarrow x = - 3, 5$

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