QJEEAdv08Ma52c9: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Ma52c9 Solution: Points E and F are given by

Question

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Points E and F are given by

Accepted Answer

Slope of line $P Q = - \sqrt{3}$

Therefore, PQ make 120 $^{\circ}$ angle with x-axis. So, side PR lies along X-axis.

Therefore, $F = (\sqrt{3}, 0)$

Now, equation CE is $\frac{x - \sqrt{3}}{\frac{- \sqrt{3}}{2}} = \frac{y - 1}{\frac{1}{2}} = 1$

$E = (\frac{\sqrt{3}}{2}, \frac{3}{2})$

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