Straight Line Question 358

Question: The line passing through $ (-1,\pi /2) $ and perpendicular to $ \sqrt{3}\sin \theta +2\cos \theta =\frac{4}{r} $ is [EAMCET 2003]

Options:

A) $ 2=\sqrt{3}r\cos \theta -2r\sin \theta $

B) $ 5=-2\sqrt{3}r\sin \theta +4r\cos \theta $

C) $ 2=\sqrt{3}r\cos \theta +2r\cos \theta $

D) $ 5=2\sqrt{3}r\sin \theta +4r\cos \theta $

Show Answer

Answer:

Correct Answer: A

Solution:

  • Perpendicular to $ \sqrt{3}\sin \theta +2\cos \theta =\frac{4}{r} $ is $ \sqrt{3}\sin ( \frac{\pi }{2}+\theta )+2\cos ( \frac{\pi }{2}+\theta )=\frac{k}{r} $ It is passing through $ (-1,\pi /2) $ $ \sqrt{3}\sin \pi +2\cos \pi =\frac{k}{-1}\Rightarrow k=2 $ \ $ \sqrt{3}\cos \theta -2\sin \theta =\frac{2}{r} $ Þ $ 2=\sqrt{3}r\cos \theta -2r\sin \theta $ .