<p>In the above problem, if \(m = \frac{\text{area of } \triangle PQR}{\text{area of } \triangle PQS}\), then \(2m\) = </p>
Step-by-Step Solution
Key Concept: Use the coordinates from problem 11 to compute the areas of both triangles using the base-height formula. The ratio simplifies directly.
<p><strong>Solution:</strong> From problem 11, we have \(P = (\sqrt{3}, \frac{3}{2})\), \(Q = (-\sqrt{3}, \frac{3}{2})\), \(R = (0, 6)\), and \(S = (0, -\frac{3}{2})\).</p><p>Area of \(\triangle PQR\): Since P and Q have the same y-coordinate, the base \(PQ = 2\sqrt{3}\) is horizontal.</p><p>Height from R to line PQ = \(|6 - \frac{3}{2}| = \frac{9}{2}\).</p><p>Area of \(\triangle PQR = \frac{1}{2} \cdot 2\sqrt{3} \cdot \frac{9}{2} = \frac{9\sqrt{3}}{2}\).</p><p>Area of \(\triangle PQS\): Height from S to line PQ = \(|\frac{3}{2} - (-\frac{3}{2})| = 3\).</p><p>Area of \(\triangle PQS = \frac{1}{2} \cdot 2\sqrt{3} \cdot 3 = 3\sqrt{3}\).</p><p>Therefore, \(m = \frac{9\sqrt{3}/2}{3\sqrt{3}} = \frac{9}{6} = \frac{3}{2}\).</p><p>Thus, \(2m = 3\).</p>
Correct Answer: 3