Q.Find 'p' and 'q' if the equation $px^2 - 8xy + 3y^2 + 14x + 2y + q = 0$ represents a pair of perpendicular lines.

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 …