<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] 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. $\square$
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