Question

Let $PQR$ be a right angled isosceles triangle, right angled at $P (2,1)$. If the equation of the line $QR$ is $2x + y = 3$, then the equation representing the pair of lines PQ and PR is

• A. $ 3x^2 - 3 y^2 + 8 xy + 20 x + 10 y + 25 = 0 $
• B. $ 3x^2 - 3 y^2 + 8 xy - 20 x - 10 y + 25 = 0 $
• C. $ 3x^2 - 3 y^2 + 8 xy + 10 x + 15 y + 25 = 0 $
• D. $ 3x^2 - 3 y^2 - 8 xy + 10 x - 15 y + 25 = 0 $

Answer 1

Answer: (B)

Let $S$ be the mid-point of $Q R$ and given $\triangle P Q R$ is an isosceles.
Therefore, $P S \perp Q R$ and $S$ is mid-point of hypotenuse, therefore $S$ is equidistant from $P, Q, R$.
$\therefore P S=Q S =R S$
Since, $ \angle P =90^{\circ} $
and $ \angle Q =\angle R$

But $\angle P+\angle Q+\angle R=180^{\circ}$
$\therefore 90^{\circ}+\angle Q+\angle R=180^{\circ}$
$\Rightarrow \angle Q=\angle R=45^{\circ}$
Now, slope of $Q R$ is $-2$. [given]
But $Q R \perp P S$.
$\therefore$ Slope of $P S$ is $1 / 2$.
Let $m$ be the slope of $P Q$.
$\therefore \tan \left(\pm 45^{\circ}\right)=\frac{m-1 / 2}{1-m(-1 / 2)}$
$\Rightarrow \pm 1=\frac{2 m-1}{2+m}$
$\Rightarrow m=3,-1 / 3$
$\therefore$ Equations of $P Q$ and $P R$ are
$y-1=3(x-2)$
and $ y-1=-\frac{1}{3}(x-2)$
or $ 3(y-1)+(x-2)=0$
Therefore, joint equation of $P Q$ and $P R$ is
${[[3(x-2)-(y-1)][(x-2)+3(y-1)]} =0$
$\Rightarrow 3(x-2)^{2}-3(y-1)^{2}+8(x-2)(y-1) =0$
$\Rightarrow 3 x^{2}-3 y^{2}+8 x y-20 x-10 y+25 =0$