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

We have
(x – 2) (x – 3) + (y – 1) (y + 4) = 0
$\Rightarrow \left(\frac{y+4}{x-2}\right)\times \left(\frac{y-1}{x-3}\right)=-1$

$\Rightarrow RP\perp RQ or \angle PRQ=\frac{\pi }{2}$
$\therefore$ The point R lies on the circle whose diameter is PQ

Now, area of $∆PQR=\frac{13}{2}$

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

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

⇒ there are two possible positions of R.