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If the straight line passing through the point $P(3,4)$ makes an angle $\frac{\pi}{6}$ with the positive direction of $X$-axis and meets the line $12 x+5 y+10=0$ at $Q$, then the length of $P Q$ is
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The correct answer is:
$\frac{132}{12 \sqrt{3}+5}$
(c) The equation of line passes through point $P(3,4)$ makes an angle $\frac{\pi}{6}$ with the positive direction of $X$-axis is
$\frac{x-3}{\cos \frac{\pi}{6}}=\frac{y-4}{\sin \frac{\pi}{6}}=r$
So, let point $Q\left(3+\frac{\sqrt{3} r}{2}, 4+\frac{r}{2}\right)$, since, the point $Q$ on the line $12 x+5 y+10=0$
So, $36+6 \sqrt{3} r+20+\frac{5}{2} r+10=0$
$\Rightarrow \quad 132+(12 \sqrt{3}+5) r=0 \Rightarrow \quad r=-\frac{132}{12 \sqrt{3}+5}$
$\therefore$ The length $P Q$ is $|r|=\frac{132}{12 \sqrt{3}+5}$
$\frac{x-3}{\cos \frac{\pi}{6}}=\frac{y-4}{\sin \frac{\pi}{6}}=r$
So, let point $Q\left(3+\frac{\sqrt{3} r}{2}, 4+\frac{r}{2}\right)$, since, the point $Q$ on the line $12 x+5 y+10=0$
So, $36+6 \sqrt{3} r+20+\frac{5}{2} r+10=0$
$\Rightarrow \quad 132+(12 \sqrt{3}+5) r=0 \Rightarrow \quad r=-\frac{132}{12 \sqrt{3}+5}$
$\therefore$ The length $P Q$ is $|r|=\frac{132}{12 \sqrt{3}+5}$
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