Straight Line Question 273
Question: Let $ P=(-1,0),Q=(0,0) $ and $ R=(3,3\sqrt{3}) $ be three point. The equation of the bisector of the angle PQR is
Options:
A) $ \frac{\sqrt{3}}{2}x+y=0 $
B) $ x+\sqrt{3y}=0 $
C) $ \sqrt{3}x+y=0 $
D) $ x+\frac{\sqrt{3}}{2}y=0 $
Correct Answer: C $ \Rightarrow \tan \theta =\sqrt{3} $ $ \Rightarrow \theta =\frac{\pi }{3}\Rightarrow \angle RQX=\frac{\pi }{3} $ $ \therefore \angle RQP=\pi -\frac{\pi }{3}=\frac{2\pi }{3}; $
Let QM bisects the $ \angle PQR, $ $ \therefore $ Slope of the line $ QM=\tan \frac{2\pi }{3}=-\sqrt{3} $ $ \therefore $ Equation of line OM is $ (y-0)=-\sqrt{3}(x-0) $ $ \Rightarrow y=-\sqrt{3}x\Rightarrow \sqrt{3}x+y=0 $
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Answer:
Solution:
BETA
