Condition for a Pair of Lines (Δ = 0)

The general second-degree equation represents a pair of straight lines iff this determinant (the discriminant of the conic) vanishes. When Δ ≠ 0, the equation represents a proper conic (circle, parabola, ellipse, or hyperbola).

For $ax^2+2hxy+by^2+2gx+2fy+c=0$ to factor as $(l_1x+m_1y+n_1)(l_2x+m_2y+n_2)=0$, the six identities in the previous derivation must be compatible.

Method — treating as quadratic in $x$:

$$ax^2 + 2(hy+g)x + (by^2+2fy+c) = 0$$

For each $y$, this gives two values of $x$ (corresponding to the two lines). For this to factor into two linear factors in both $x$ and $y$, the discriminant of the quadratic in $x$ must itself be a perfect square in $y$:

$$\Delta_x = (hy+g)^2 - a(by^2+2fy+c)$$ $$= h^2y^2+2hgy+g^2 - aby^2-2afy-ac$$ $$= (h^2-ab)y^2 + 2(hg-af)y + (g^2-ac)$$

For this to be a perfect square: discriminant of this quadratic in $y$ $= 0$:

$$4(hg-af)^2 - 4(h^2-ab)(g^2-ac) = 0$$

Expanding:

$$h^2g^2 - 2afghg + a^2f^2 - h^2g^2 + ach^2 + abg^2 - a^2bc = 0$$ $$-2afgh + a^2f^2 + ach^2 + abg^2 - a^2bc = 0$$

Dividing by $a$:

$$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$$

This is $\Delta = 0$, where $\Delta$ is the determinant of the matrix $\begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix}$.