Explanation

(b)
Let M be mid point of $QR$. As $PQR$ is an isosceles triangle, $PM\bot QR$. Slope of $QR$ is $- 2$
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⇒ Slope of $PM$ is 1/2. Since, $\mathrm{\Delta}QPR$ is a right angled triangle, $Q,\ P,\ R$ lie on a circle with centre at $M$.
$\therefore MQ = MP = MR$
$\Rightarrow \ \angle QPM = 45{^\circ}$
Let $m$ be slope of $PQ$. Thus,
$\pm \tan{45{^\circ}} = 1 = \frac{m - \frac{1}{2}}{1 + \frac{m}{2}}$
$\Rightarrow \ \pm 1 = \frac{2m - 1}{2 + m}$
$\Rightarrow \ \pm (2 + m) = 2m - 1 \Rightarrow m = 3,\ - \frac{1}{3}$
$\therefore$ Equation of $PQ$ and $PR$ are
$y - 1 = 3(x - 2)$
and $y - 1 = - \frac{1}{3}(x - 2)$
$\Rightarrow 3(y - 1) + (x - 2) = 0$
Thus, joint equation of $PQ$ and $PR$ is
$\left\lbrack 3(x - 2) - (y - 1) \right\rbrack\left\lbrack (x - 2) + 3(y - 1) \right\rbrack = 0$
$\Rightarrow 3({x - 2)}^{2} - 3(y - 1)^{2} + 8(x - 2)(y - 1) = 0$
$\Rightarrow \ {3x}^{2} - {3y}^{2} + 8xy - 20x - 10y + 25 = 0$