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Q.
P(3, 1), Q (6, 5) and R (x, y) are three points such that the angle RPQ is a right angle and the area of $\Delta$RQP = 7, then the number of such points R is
Straight Lines
Solution:
Since $\angle RPQ = \frac{\pi}{2}$, therefore
Slope of RP $\times$ slope of PQ = -1
$\Rightarrow \frac{y-1}{x-3} \times\frac{5-1}{6-3} = - 1 \Rightarrow 3x +4y =13$ ...(i)
Also, area of $ \Delta RPQ = 7 \Rightarrow \begin{vmatrix}x&y&1\\ 3 &1&1\\ 6&5&1\end{vmatrix} = \pm7$
$ \Rightarrow 3y - 4x = 5 or 3y - 4x = -23$ ...(ii)
Solving (i) and (ii), we get two points .