Straight Line Question 369

Question: The equation of the lines on which the perpendiculars from the origin make $ 30^{o} $ angle with x?axis and which form a triangle of area $ \frac{50}{\sqrt{3}} $ with axes, are

Options:

A) $ x+\sqrt{3}y\pm 10=0 $

B) $ \sqrt{3}x+y\pm 10=0 $

C) $ x\pm \sqrt{3}y-10=0 $

D)None of these

Show Answer

Answer:

Correct Answer: B

Solution:

  • Let p be the length of the perpendicular from the origin on the given line. Then its equation in normal form is $ x\cos 30^{o}+y\sin 30^{o}=p $ or $ \sqrt{3}x+y=2p $ This meets the coordinate axes at $ A( \frac{2p}{\sqrt{3}},0 ) $ and $ B(0,2p) $ . \ Area of $ \Delta OAB=\frac{1}{2}( \frac{2p}{\sqrt{3}} )2p=\frac{2p^{2}}{\sqrt{3}} $ By hypothesis $ \frac{2p^{2}}{\sqrt{3}}=\frac{50}{\sqrt{3}}\Rightarrow p=\pm 5 $ . Hence the lines are $ \sqrt{3}x+y\pm 10=0 $ .