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$P (x, y)$, $Q (-2, – 3)$ and $R (2, 3)$ are the vertices of a right triangle PQR right angled at P. Find the relationship between $x$ and $y$. Hence, find all possible values of $x$ for which $y = 2$.
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In $\triangle PQR$, $\angle P = 90^\circ$
$PQ^2 + PR^2 = QR^2$
$\Rightarrow (x + 2)^2 + (y + 3)^2 + (x – 2)^2 + (y – 3)^2 = 4^2 + 6^2$
$\Rightarrow x^2 + 4x + 4 + y^2 + 6y + 9 + x^2 – 4x + 4 + y^2 – 6y + 9 = 52$
gives, $x^2 + y^2 = 13$
Now for $y = 2, x = \pm 3$
$PQ^2 + PR^2 = QR^2$
$\Rightarrow (x + 2)^2 + (y + 3)^2 + (x – 2)^2 + (y – 3)^2 = 4^2 + 6^2$
$\Rightarrow x^2 + 4x + 4 + y^2 + 6y + 9 + x^2 – 4x + 4 + y^2 – 6y + 9 = 52$
gives, $x^2 + y^2 = 13$
Now for $y = 2, x = \pm 3$