A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{2}\]
D) \[0\]
Correct Answer: C
Solution :
Key Idea: Angle between the pair of lines\[{{x}^{2}}+2xy-{{y}^{2}}=0\]is \[\theta ={{\tan }^{-1}}\frac{\sqrt{{{h}^{2}}-ab}}{a+b}\] Given pair of lines is \[{{x}^{2}}+2xy-{{y}^{2}}=0\] Here, \[a=1,\,\,h=1,\,\,b=-1\] \[\therefore \] \[\theta ={{\tan }^{-1}}\frac{\sqrt{1+1}}{1-1}={{\tan }^{-1}}(\infty )\] Note: If pair of lines is parallel, then \[{{h}^{2}}-ab=0\]
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