<s>[INST] 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. 
 [/INST] $$( A^2 - 3 B^2 ) x^2 + 8 A B x y + ( B^2 - 3 A^2 ) y^2 = 0 $$
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
$$ y = m_1 x , y = m_2 x ; \,( m_1, m_2) =   \frac {4\, A B \pm \sqrt{3}(A^2 +B^2 )}{ (A^2 -B^2 )} $$ 

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