Question: Angle between the lines $ 2x-y-15=0 $ and $ 3x+y+4=0 $ is[RPET 2003]
Options:
A) $ 90^{o} $
B) $ 45^{o} $
C) $ 180^{o} $
D) $ 60^{o} $
Show Answer
Answer:
Correct Answer: A
Solution:
- Lines are $ p=| \frac{-k}{\sqrt{{{\sec }^{2}}\alpha +cose{c^{2}}\alpha }} | $ …..(i) and $ 3x+y+4=0 $……(ii) Here, $ m_1=2,m_2=-3 $ If angle between them is $ \theta $ , then $ \tan \theta =| \frac{m_1-m_2}{1+m_1m_2} | $ $ =| \frac{2+3}{1-6} |=| \frac{5}{-5} | $ = 1 $ \tan \theta =\tan \frac{\pi }{4} $
Þ $ \theta =\frac{\pi }{4}=45{}^\circ . $