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

Answer 1

Answer: (B)

Let $S$ be the mid-point of $QR$ and given $△PQR$ is an isosceles.
Therefore, $PS\perp QR$ and $S$ is mid-point of hypotenuse, therefore $S$ is equidistant from $P,Q,R$.
$\therefore PS=QS=RS$
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 $QR$ is $-2$. [given]
But $QR\perp PS$.
$\therefore$ Slope of $PS$ is $1/2$.
Let $m$ be the slope of $PQ$.
$\therefore \tan \left(\pm 45^{\circ }\right)=\frac{m-1/2}{1-m(-1/2)}$
$\Rightarrow \pm 1=\frac{2m-1}{2+m}$
$\Rightarrow m=3,-1/3$
$\therefore$ Equations of $PQ$ and $PR$ 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 $PQ$ and $PR$ is
$[[3(x-2)-(y-1)][(x-2)+3(y-1)]=0$
$\Rightarrow 3(x-2{)}^2-3(y-1{)}^2+8(x-2)(y-1)=0$
$\Rightarrow 3x^2-3y^2+8xy-20x-10y+25=0$