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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

  1. A \(6 \sqrt{2}\)
  2. B \(9 \sqrt{2}\)
  3. C \(12 \sqrt{2}\)
  4. D \(18 \sqrt{2}\)
Verified Solution

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…