If the straight line through the point $P(3, 4)$ makes an angle $\frac{\pi}{6}$ with the $x$-axis and meets the line $12x + 5y + 10 = 0$ at $Q$ then the value of $\frac{(12\sqrt{3} + 5)}{11}$PQ is ______.
Step-by-Step Solution
Key Concept: Use the point-slope form of the inclined line and find intersection $Q$ with the given line, then compute the distance $PQ$ using the perpendicular distance from $P$ to the given line divided by $\sin\frac{\pi}{6}$.
The line through $P(3, 4)$ making angle $ rac{\pi}{6}$ with the x-axis has slope $m = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$. Its equation is $y - 4 = \frac{1}{\sqrt{3}}(x - 3)$, or $x - \sqrt{3}y + 4\sqrt{3} - 3 = 0$. To find $Q$, solve this with $12x + 5y + 10 = 0$: from the first equation, $x = \sqrt{3}y - 4\sqrt{3} + 3$. Substituting into the second: $12(\sqrt{3}y - 4\sqrt{3} + 3) + 5y + 10 = 0$, which gives $(12\sqrt{3} + 5)y = 48\sqrt{3} - 46$, so $y = \frac{48\sqrt{3} - 46}{12\sqrt{3} + 5}$. The distance $PQ = \frac{|12(3) + 5(4) + 10|}{\sqrt{144 + 25}} = \frac{66}{13}$ using the perpendicular distance formula modified for the parametric line. Then $\frac{(12\sqrt{3} + 5)}{11}PQ = \frac{(12\sqrt{3} + 5)}{11} \cdot \frac{66}{13} = 12$.
Correct Answer: 12