• # question_answer 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 length $PQ$is A)            $\frac{132}{12\sqrt{3}+5}$     B)            $\frac{132}{12\sqrt{3}-5}$ C)            $\frac{132}{5\sqrt{3}+12}$     D)            $\frac{132}{5\sqrt{3}-12}$

The equation of any line passing through the given point P(3, 4) and making an angle $\frac{\pi }{6}$with x-axis is $\frac{x-3}{\cos {{30}^{o}}}=\frac{y-4}{\sin {{30}^{o}}}=r$   (say)           ......(i)                    Where ?r? represents the distance of any point Q on this line from the given point P (3, 4).                    The coordinates (x, y) of any point Q on line (i) are $(3+r\cos {{30}^{o}},4+r\sin {{30}^{o}})\,\,\text{ }i.e.,\,\text{ }\left( 3+\frac{r\sqrt{3}}{2},4+\frac{r}{2} \right)$                    If the point lies on the line $12x+5y+10=0$, then                    $12\left( 3+\frac{r\sqrt{3}}{2} \right)+5\left( 4+\frac{r}{2} \right)+10=0\Rightarrow r=\frac{132}{12\sqrt{3}+5}$.