Question: 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
Options:
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} $
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Answer:
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})i.e.,( 3+\frac{r\sqrt{3}}{2},4+\frac{r}{2} ) $ If the point lies on the line $ 12x+5y+10=0 $ , then $ 12( 3+\frac{r\sqrt{3}}{2} )+5( 4+\frac{r}{2} )+10=0\Rightarrow r=\frac{132}{12\sqrt{3}+5} $ .
BETA
