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 132, then the number of possible positions of R are

We have(x – 2) (x – 3) + (y – 1) (y + 4) = 0⇒y+4x-2×y-1x-3=-1⇒RP⊥RQ or ∠PRQ=π2∴ The point R lies on the circle whose diameter is PQNow, area of ∆PQR=132⇒12×26×altitude=132⇒altitude=262=radius⇒ there are two possible positions of R.