Area of triangle bounded by line and degenerate "crossed lines" conic The question is 

Show that the two lines given by 
  $$(A^2 - 3B^2)x^2 + 8ABxy +(B^2 - 3A^2)y^2=0$$
  and the line given by $$Ax+By+C=0$$ determine an equilateral triangle of area $$\frac{C^2}{\sqrt{3}\;(A^2+B^2)}$$

I tried  factorizing the pair of straight lines into two straight lines. But that seemed to be not really simple. And I doubt I was proceeding rightly. It would be great if anyone helps. 
 A: A simple way: let us apply the similarity transform $u=Ax+By,v=Bx-Ay$.
The line equation becomes $$Ax+By+C=u+C=0$$ and the line pair equation
$$(A^2-3B^2)x^2+8ABxy+(B^2-3A^2)y^2=u^2-3v^2=(u-\sqrt3v)(u+\sqrt3v)=0.$$
The pairwise intersections are $(0,0),(-C,\dfrac C{\sqrt3}),(-C,-\dfrac C{\sqrt3})$, thus the three squared side lengths are $\dfrac43C^2$ and the triangle is equilateral.
Correcting by the similarity ratio,
$$Area=\frac{\sqrt3}4\left(\frac43C^2\right)\left(\frac1{\sqrt{A^2+B^2}}\right)^2=\frac{C^2}{\sqrt3(A^2+B^2)}.$$
A: The crossed lines defined by
 $$( A^2 - 3 B^2 ) x^2 + 8 A B x y + ( B^2 - 3 A^2 ) y^2 = 0 \qquad(\star)$$
pass through the origin, so one of our vertices is $O(0,0)$.
Convenient manipulation allows us to re-write $(\star)$ thusly:
$$x^2 + y^2 = \frac{4( Ax + B y )^2}{3(A^2+B^2)} \qquad(\star\star)$$
If $P(x,y)$ satisfies $(\star)$ (that is, $(\star\star)$) and also $A x + B y + C = 0$, then we have
$$x^2 + y^2 = \frac{4C^2}{3(A^2+B^2)} \quad\text{so that}\quad |\overline{OP}| = \frac{2|C|}{\sqrt{3}\;\sqrt{A^2+B^2}}$$ 
But $d := |C|/\sqrt{A^2+B^2}$ is precisely the distance from $O$ to the line defined by $Ax+By+C=0$. Consequently, if $P_1$ and $P_2$ are the other two vertices of our triangle, then 
$$|\overline{OP_1}| = |\overline{OP_2}| = \frac{2}{\sqrt{3}}d$$
That is, $\triangle OP_1P_2$ is isosceles. Moreover, the ratio of its base altitude ($d$) to its leg makes it equilateral, and we can write
$$|\triangle OP_1P_2| = \frac{1}{2} \;d\;|\overline{OP_1}| = \frac{1}{2} d^2 \frac{2}{\sqrt{3}} = \frac{C^2}{\sqrt{3}\;(A^2+B^2)}$$
as desired. $\square$
A: $$( A^2 - 3 B^2 ) x^2 + 8 A B x y + ( B^2 - 3 A^2 ) y^2 = 0 $$
are a pair of straight lines through the the origin making angle $ \pi/3 $ or  $2 \pi/3 $ with each other at the origin, just like two sides along diameter of circum-circle of a regular hexagon.
They are factorisable into two straight lines as it is easy to verify that for two cases in particular :
$$ A = 1, B=0 \rightarrow x^2 - 3 y^2 = 0 ; A = 0, B=1 \rightarrow y^2 - 3 x^2 = 0 $$
and, in general,
$$ (A^2-3 B^2) x^2 - 8 A B x y + ( B^2 - 3 A^2 ) y^2  = 0 $$ has slopes
$$ y = m_1 x , y = m_2 x ; \,( m_1, m_2) =   \frac {4\, A B \pm \sqrt{3}(A^2 +B^2 )}{ (A^2 -B^2 )} $$ 

