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If $O$ is the origin and $P, Q$ are points on the line $3 x+4 y+$ $15=0$ such that $\mathrm{OP}=\mathrm{OQ}=9$, then the area of $\triangle \mathrm{OPQ}$ is
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Verified Answer
The correct answer is:
$18 \sqrt{2}$
Let $M$ be the foot of perpendicular drawn from $O(0,0)$ on line $3 x+4 y+15=0$
Hence $O M=\left|\frac{3 \times 0+4 \times 0+15}{\sqrt{3^2+4^2}}\right|=3$
Now since $\triangle O M Q$,
$$
M Q=\sqrt{81-9}=6 \sqrt{2}
$$
Area of $\triangle O P Q=2[$ Area of $\triangle O M Q]$
$$
\begin{aligned}
& =2\left[\frac{1}{2}(O M)(M Q)\right] \\
& =3 \times 6 \sqrt{2}=18 \sqrt{2}
\end{aligned}
$$
Hence $O M=\left|\frac{3 \times 0+4 \times 0+15}{\sqrt{3^2+4^2}}\right|=3$
Now since $\triangle O M Q$,
$$
M Q=\sqrt{81-9}=6 \sqrt{2}
$$
Area of $\triangle O P Q=2[$ Area of $\triangle O M Q]$
$$
\begin{aligned}
& =2\left[\frac{1}{2}(O M)(M Q)\right] \\
& =3 \times 6 \sqrt{2}=18 \sqrt{2}
\end{aligned}
$$
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