AP EAMCET · Maths · Straight Lines
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
- A \(6 \sqrt{2}\)
- B \(9 \sqrt{2}\)
- C \(12 \sqrt{2}\)
- D \(18 \sqrt{2}\)
Answer & Solution
Correct Answer
(D) \(18 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
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…
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