To prove triangle is equilateral> given one equation and one combined equation. show that the lines $ x^2 - 4xy +y^2 $ and $x +y=3 $ form an equilateral trinagle . Also Find Area.
Here is what I have tried:
$ l1 =x+y=3 $
The combined equation is:
$ x^2 - 4xy + y^2 $
compare it with general formula :
$ ax^2 +2hxy+y^2  $
What to do next ? 
I could find the slope but how will that help ?

Put $y = mx$ in the joint equation. That will produce auxiliary equation $$m^2 - 4m + 1 = 0$$
Solving this will give $$m = \dfrac{4 \pm \sqrt{4^2 - 4}}{2} = \dfrac{4 \pm 2\sqrt3}{2} = 2\pm \sqrt{3}$$
Let $m_1 = 2+\sqrt3, m_2 = 2-\sqrt3$
Gradient of $x+y = 1$ is $-1$, let $m_3 = -1$.
Use $\tan\theta = \left|\dfrac{m_a-m_b}{1+m_am_b}\right|$ for all $3$ pairs and prove that it is $\tan \dfrac\pi3 = \sqrt3$
$$\tan\theta_{m_1m_2} = \left|\dfrac{(2+\sqrt3)-(2-\sqrt3)}{1+(2+\sqrt3)(2-\sqrt3)}\right| = \left|\dfrac{2\sqrt3}{1+4-3}\right| = \sqrt3$$
$$\tan\theta_{m_1m_3} = \left|\dfrac{2+\sqrt3+1}{1-2-\sqrt3}\right| = \left|\dfrac{\sqrt3(1+\sqrt3)}{-(1+\sqrt3)} \right| = \sqrt3$$
$\theta_{m_2m_3}$ can be calculated similarly.
