Q.If the lines given by $ax^2 + 2hxy + by^2 = 0$ form an equilateral triangle with the line $lx + my = 1$ then show that $(3a+b)(a+3b) = 4h^2$.

The combined equation of any two lines (not necessarily through the origin) always has the fully general second-degree shape $ax^2+2hxy+by^2+2gx+2fy+c=0$, but the converse fails -- equations like $x^2+y^2=25$ have the same shape yet describe a circle, not two lines. A general second-degree equation genuinely represents a pair of lines only when both $abc+2fgh-af^2-bg^2-ch^2=0$ (equivalently, the $3\times3$ determinant with $a,b,c$ on the diagonal and $h,g,f$ off it vanishes) and $h^2-ab\geq0$ hold together; in practice the safest way to confirm this and find the two lines is to factor the quadratic part and match the remaining linear/constant terms by comparison. Once confirmed, the pair is parallel to the lines of its homogeneous part $ax^2+2hxy+by^2=0$, so the same angle formula, perpendicularity condition $a+b=0$, and parallel condition $h^2-ab=0$ carry over unchanged, and (when $h^2-ab>0$) the two lines meet at $\left(\df …