Pair Of Straight Lines Question 42
Question: The lines $ {{(lx+my)}^{2}}-3{{(mx-ly)}^{2}}=0 $ and $ lx+my+n=0 $ form
Options:
A) An isosceles triangle
B) A right angled triangle
C) An equilateral triangle
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Lines are $ [(l+\sqrt{3}m)x+(m-\sqrt{3}l)y][(l-\sqrt{3}m)x+(m+\sqrt{3}l)y]=0 $ and $ L_3=lx+my+n=0 $ .
L1 and L2 are above two lines. $ S_1=-\frac{(l+\sqrt{3}m)}{(m-\sqrt{3}l)},\ \ \ S_2=-\frac{(l-\sqrt{3}m)}{(m+\sqrt{3}l)},\ \ \ S_3=-\frac{l}{m} $ (where $ S_1 $ , $ S_2 $ and $ S_3 $ are slopes of the lines) $ {\theta _{13}}={{\tan }^{-1}}[ \frac{-( \frac{l+\sqrt{3}m}{m-\sqrt{3}l} )+\frac{l}{m}}{1+( \frac{l+\sqrt{3}m}{m-\sqrt{3}l} )\frac{l}{m}} ] $
$ ={{\tan }^{-1}}( \frac{-\sqrt{3}m^{2}-\sqrt{3}l^{2}}{l^{2}+m^{2}} )=60{}^\circ $
$ {\theta _{23}}={{\tan }^{-1}}[ \frac{-( \frac{l-\sqrt{3}m}{m+\sqrt{3}l} )+\frac{l}{m}}{1+( \frac{l-\sqrt{3}m}{m+\sqrt{3}l} )( \frac{l}{m} )} ] $
$ ={{\tan }^{-1}}( \frac{\sqrt{3}m^{2}+\sqrt{3}l^{2}}{m^{2}+l^{2}} )={{\tan }^{-1}}(\sqrt{3})=60{}^\circ $ Hence, triangle is equilateral
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