Question

The figure formed by the lines $x^2+4xy+y^2=0$ and x - y = 4, is [Roorkee 1980]

• A. A right angled triangle
• B. An isosceles triangle
• C. An equilateral triangle
• D. None of these

Answer 1

The correct option is C

An equilateral triangle

$S_1=\frac{1}{-2+\sqrt{4-1}}=\frac{1}{-2+\sqrt{3}}=-(\sqrt{3}+2)$

$S_1=\frac{1}{-2-\sqrt{4-1}}=\frac{1}{-2-\sqrt{3}}=(\sqrt{3}-2)$ and $S_3=1$.

${\theta }_{13}=ta{n}^{-1}\left|\frac{-(\sqrt{3}+2)-1}{1-(\sqrt{3}+2)}\right|=ta{n}^{-1}\left|\frac{-(\sqrt{3}+3)}{-(\sqrt{3}+1)}\right|$

= ${\tan }^{-1}\left(\sqrt{3}\right)={60}^{\circ }$.