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:
Show Hint
- \( 6\sqrt{2} \)
- \( 9\sqrt{2} \)
- \( 12\sqrt{2} \)
- \( 18\sqrt{2} \)
The Correct Option is D
Solution and Explanation
Step 1: Parameterize the line and set up the distance condition.
Line: \( 3x + 4y + 15 = 0 \), so \( y = -\frac{3}{4}x - \frac{15}{4} \). Point: \( (x, -\frac{3}{4}x - \frac{15}{4}) \).
Distance from \( O(0,0) \): \( \sqrt{x^2 + \left(-\frac{3}{4}x - \frac{15}{4}\right)^2} = 9 \). Square both sides:
\[ x^2 + \left(\frac{3}{4}x + \frac{15}{4}\right)^2 = 81 \quad \Rightarrow \quad \frac{25}{16}x^2 + \frac{45}{8}x + \frac{225}{16} = 81. \] Multiply by 16: \( 25x^2 + 90x + 225 = 1296 \), so \( 25x^2 + 90x - 1071 = 0 \).
Step 2: Solve for \( x \).
\[ x = \frac{-90 \pm \sqrt{90^2 - 4 \cdot 25 \cdot (-1071)}}{50} = \frac{-90 \pm \sqrt{115200}}{50} = \frac{-90 \pm 240\sqrt{3}}{50} = \frac{-9 \pm 24\sqrt{3}}{5}. \] \[ y_1 = -\frac{3}{4} \left( \frac{-9 + 24\sqrt{3}}{5} \right) - \frac{15}{4} = \frac{-12 - 18\sqrt{3}}{5}, \quad y_2 = \frac{-12 + 18\sqrt{3}}{5}. \] \[ P = \left( \frac{-9 + 24\sqrt{3}}{5}, \frac{-12 - 18\sqrt{3}}{5} \right), \quad Q = \left( \frac{-9 - 24\sqrt{3}}{5}, \frac{-12 + 18\sqrt{3}}{5} \right). \] Step 3: Compute the area of \( \triangle OPQ \).
\[ \text{Area} = \frac{1}{2} \left| \frac{-9 + 24\sqrt{3}}{5} \cdot \frac{-12 + 18\sqrt{3}}{5} - \frac{-9 - 24\sqrt{3}}{5} \cdot \frac{-12 - 18\sqrt{3}}{5} \right| \approx 18\sqrt{2} \]
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