Question

Let $P(2,-4)$ and $Q(3,1)$ be two given points. Let $R(x,y)$ be a point such that $(x-2)(x-3)+(y-1)(y+4)=0$. If area of $△PQR$ is $\frac{13}{2}$, then the number of possible positions of $R$ are

• A. $2$
• B. $3$
• C. $4$
• D. None of these

Answer 1

Answer: (A)

$2$

We have $(x-2)(x-3)+(y-1)(y+4)=0$

$\Rightarrow \left(\frac{y+4}{x-2}\right)\times \left(\frac{y-1}{x-3}\right)=-1$

$\Rightarrow RP\perp RQ$ or $\angle PQR=\frac{\pi }{2}$

$\therefore$ The point $R$ lies on the circle whose diameter is $PQ$

Now, area of $△PQR=\frac{13}{2}$

$\Rightarrow \frac{1}{2}\times \sqrt{26}\times ($altitude$)=\frac{13}{2}$

$\Rightarrow$ altitude $=\frac{\sqrt{26}}{2}=$ radius

$\Rightarrow$ There are two possible positions of $R.$