Pair Of Straight Lines Question 124
Question: The figure formed by the lines $ x^{2}+4xy+y^{2}=0 $ and $ x-y=4, $ is
[Roorkee 1980]
Options:
A) A right angled triangle
B) An isosceles triangle
C) An equilateral triangle
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ S_1=\frac{1}{-2+\sqrt{4-1}}=\frac{1}{-2+\sqrt{3}}=\frac{-\sqrt{3}-2}{(-2+\sqrt{3})(-2-\sqrt{3})} $
$ S_2=\frac{1}{-2-\sqrt{4-1}}=\frac{1}{-2-\sqrt{3}}= \frac{\sqrt{3}-2}{(-2-\sqrt{3})(\sqrt{3}-2)} $
and $ S_3=1 $ . $ {\theta _{13}}={{\tan }^{-1}}| \frac{-(\sqrt{3}+2)-1}{1-(\sqrt{3}+2)} |={{\tan }^{-1}}| \frac{-(\sqrt{3}+3)}{-(\sqrt{3}+1)} | $
$ ={{\tan }^{-1}}(\sqrt{3})=60{}^\circ $ . $ {\theta _{23}}={{\tan }^{-1}}\left| \frac{\sqrt{3}-2-1}{1+\sqrt{3}-2} \right|={{\tan }^{-1}}\left| \frac{\sqrt{3}-3}{\sqrt{3}-1} \right| $
$ ={{\tan }^{-1}}(\sqrt{3})=60{}^\circ $ .
BETA
