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

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 $$132$$, then the number of possible positions of R are

Infinity Learn · Accepted Answer

We $$have(x$$ – $$2)$$ (x$$ – $$3) $$+$$ (y$$ – $$1) (y$$ $$+$$ $$4) $$=$$ $$0$$⇒$$y+4x-2$$×$$y-1x-3=-1$$⇒RP⊥RQ or ∠$$PRQ=$$π$$2$$∴ 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.

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

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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

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

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