First slide

Equation of line in Straight lines

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

Moderate

A

2
B

3
C

4
D

none of these

Solution

We have
(x – 2) (x – 3) + (y – 1) (y + 4) = 0
$\Rightarrow (\frac{y + 4}{x - 2}) \times (\frac{y - 1}{x - 3}) = - 1$
$\Rightarrow R P ⊥ R Q o r ∠ P R Q = \frac{π}{2}$
$∴$ The point R lies on the circle whose diameter is PQ
Now, area of $∆ P Q R = \frac{13}{2}$
$\Rightarrow \frac{1}{2} \times \sqrt{26} \times (a l t i t u d e) = \frac{13}{2}$
$\Rightarrow a l t i t u d e = \frac{\sqrt{26}}{2} = r a d i u s$
⇒ there are two possible positions of R.

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