P(3,1),Q(6,5) and R(x,y) are three points suc | vedclass

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(D) Since $\angle PRQ = 90^{\circ}$,the point $R$ must lie on a circle with $PQ$ as its diameter.The length of the diameter $PQ = \sqrt{(6-3)^2 + (5-1

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$4$

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(D) Since $\angle PRQ = 90^{\circ}$,the point $R$ must lie on a circle with $PQ$ as its diameter.The length of the diameter $PQ = \sqrt{(6-3)^2 + (5-1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5$.The radius of this circle is $r = \frac{5}{2} = 2.5$.The area of $\Delta RPQ = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = 5$.$\frac{1}{2} 	imes PQ 	imes h = 5 \implies \frac{1}{2} 	imes 5 	imes h = 5 \implies h = 2$.Here,$h$ is the perpendicular distance of point $R$ from the line segment $PQ$.Since the maximum height of any point on the circle from the diameter $PQ$ is the radius $r = 2.5$,and we require $h = 2$,there exist two points on the upper semi-circle and two points on the lower semi-circle that satisfy this condition.Thus,there are $4$ such points $R$.

$P(3,1)$,$Q(6,5)$ and $R(x,y)$ are three points such that the angle $\angle PRQ$ is a right angle and the area of the $\Delta RPQ = 5$. Then the number of such points $R$ is

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