A) \[\frac{132}{12\sqrt{3}+5}\]
B) \[\frac{132}{12\sqrt{3}-5}\]
C) \[\frac{132}{5\sqrt{3}+12}\]
D) \[\frac{132}{5\sqrt{3}-12}\]
Correct Answer: A
Solution :
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 \[12\left( 3+\frac{r\sqrt{3}}{2} \right)+5\left( 4+\frac{r}{2} \right)+10=0\Rightarrow r=\frac{132}{12\sqrt{3}+5}\].
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