Question

Show that the lines $x^2+4xy+y^2=0$ and $x-y=4$ form an equilateral triangle. Also find the area of the triangle.

Answer 1

Given pair of lines are $x^2+4xy+y^2=0$

$⟹{\left(\frac{y}{x}\right)}^2+4\frac{y}{x}+1=0$

$⟹\frac{y}{x}=\frac{-4\pm 2\sqrt{3}}{2}=-2\pm \sqrt{3}$,

$\therefore$ The lines $y=(-2+\sqrt{3})x$ and $y=(-2-\sqrt{3})x$ and $x-y=4$ forms an equilateral triangle

Clearly the pair of lines $x^2+4xy+y^2=0$ intersect at origin,

The perpendicular distance form origin to $x-y=4$ is the height of the equilateral triangle

$h=2\sqrt{2}$

$\therefore$ Area of the triangle is $\frac{h^2}{\sqrt{3}}=\frac{8}{\sqrt{3}}$