If the points P(3, 1) and Q(6, 5) form a tria | vedclass

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(D) Given that $\angle PRQ = \frac{\pi}{2}$,the point $R$ must lie on a circle with $PQ$ as its diameter.Length of $PQ = \sqrt{(6-3)^2 + (5-1)^2} = \s

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$4$

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(D) Given that $\angle PRQ = \frac{\pi}{2}$,the point $R$ must lie on a circle with $PQ$ as its diameter.Length of $PQ = \sqrt{(6-3)^2 + (5-1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5$.The area of $\Delta PQR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes PQ 	imes h = 6$.$\frac{1}{2} 	imes 5 	imes h = 6 \implies h = \frac{12}{5} = 2.4$.The maximum height of the triangle is the radius of the circle,which is $r = \frac{PQ}{2} = \frac{5}{2} = 2.5$.Since $h = 2.4 Each line intersects the circle at two distinct points.Therefore,there are $2 + 2 = 4$ possible positions for point $R$.

If the points $P(3, 1)$ and $Q(6, 5)$ form a triangle with a third point $R(x, y)$ such that the area of $\Delta PQR$ is $6$ square units and $\angle PRQ = \frac{\pi}{2}$,then the number of possible positions for point $R$ is:

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