If O is the origin and P, Q are points on the | vedclass

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(D) Let $M$ be the foot of the perpendicular drawn from $O(0,0)$ to the line $3x + 4y + 15 = 0$.Hence, the length of the perpendicular $OM = \left| \f

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$18$

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(D) Let $M$ be the foot of the perpendicular drawn from $O(0,0)$ to the line $3x + 4y + 15 = 0$.Hence, the length of the perpendicular $OM = \left| \frac{3(0) + 4(0) + 15}{\sqrt{3^2 + 4^2}} ight| = \frac{15}{5} = 3$.In the right-angled triangle $	riangle OMQ$, by the Pythagorean theorem:$MQ = \sqrt{OQ^2 - OM^2} = \sqrt{9^2 - 3^2} = \sqrt{81 - 9} = \sqrt{72} = 6 \sqrt{2}$.The area of $	riangle OPQ = 2 	imes (	ext{Area of } 	riangle OMQ)$.Area of $	riangle OPQ = 2 	imes \left( \frac{1}{2} 	imes OM 	imes MQ ight) = OM 	imes MQ$.Area of $	riangle OPQ = 3 	imes 6 \sqrt{2} = 18 \sqrt{2}$.

If $O$ is the origin and $P, Q$ are points on the line $3x + 4y + 15 = 0$ such that $OP = OQ = 9$, then the area of $\triangle OPQ$ is (in $\sqrt{2}$)

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