Q.Show that $2x^2 + 3xy - 2y^2 + x + 7y - 3 = 0$ represents a pair of straight lines, and find the separate equations of the two lines.

Not every pair of straight lines passes through the origin, so we need a general second-degree equation, $ax^2+2hxy+by^2+2gx+2fy+c=0$, to describe an arbitrary pair anywhere in the plane. But this general form usually describes a genuine conic (an ellipse, parabola, or hyperbola), not two straight lines — so we need a precise algebraic test for when it degenerates into a pair of lines.

Treat the equation as a quadratic in $x$: $ax^2+2(hy+g)x+(by^2+2fy+c)=0$. For the two roots for $x$ to be linear expressions in $y$ (rather than genuinely involving a square root of a non-perfect-square in $y$), the discriminant $(hy+g)^2-a(by^2+2fy+c)$, when expanded as $(h^2-ab)y^2+2(hg-af)y+(g^2-ac)$, must itself be a perfect square in $y$ — which happens exactly when its own discriminant vanishes: $(hg-af)^2-(h^2-ab)(g^2-ac)=0$. Expanding and simplifying this (a lengthy but standard piece of algebra) reduces it to the compact, symmetric condition

equivalently the vanishing of the determinant $\begin{vmatrix}a&h&g\\h&b&f\\g&f&c\end{vmatrix}=0$. This is the master algebraic test: whenever it holds, the second-degree equation factors into two linear expressions.

However, satisfying this determinant condition alone only guarantees the equation factors — it does not by itself guarantee the two resulting lines are real. Exactly as in the purely homogeneous case, we additionally need $h^2\geq ab$ (viewing the equation as a quadratic in $x/y$-type ratios of the leading terms), and by parallel reasoning eliminating $x$ or $y$ first, $g^2\geq ac$ and $f^2\geq bc$ as well — three inequalities that together certify that no part of the factorisation secretly involves an imaginary coefficient.

As a worked check: $(x+2y-1)(2x-y+3)=0$ expands to $2x^2+3xy-2y^2+x+7y-3=0$, giving $a=2,h=\tfrac32,b=-2,g=\tfrac12,f=\tfrac72,c=-3$. Substituting into $abc+2fgh-af^2-bg^2-ch^2$ gives exactly zero, confirming algebraically that this equation is indeed a genuine pair of straight lines — matching its known factorisation into $x+2y-1=0$ and $2x-y+3=0$. …