The triangle formed by $x^2-4 x y+y^2=0$ and $x+y+4 \sqrt{6}=0$ is

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Aicharya · Accepted Answer

$x^2-4 x y+y^2=0$ $\Rightarrow y^2-2 \cdot 2 x y+4 x^2=3 x^2$ $\begin{array}{ll}\Rightarrow & (y-2 x)^2=3 x^2 \\ \Rightarrow & y-2 x= \pm \sqrt{3} x\end{array}$ $y=(2 \pm \sqrt{3}) x$

$\therefore$ The given triangle is an equilateral triangle.

The triangle formed by $x^2-4 x y+y^2=0$ and $x+y+4 \sqrt{6}=0$ is

Solution