Condition for General Second Degree to be a Pair

i.e. abc + 2fgh − af² − bg² − ch² = 0. This is the necessary and sufficient condition.

The general second degree equation:

$$S \equiv ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$

represents a pair of lines if it factors as $(l_1x+m_1y+n_1)(l_2x+m_2y+n_2) = 0$.

Derivation

Expanding the product and comparing coefficients:

$$a = l_1l_2,\quad b = m_1m_2,\quad c = n_1n_2$$ $$2h = l_1m_2+l_2m_1,\quad 2g = l_1n_2+l_2n_1,\quad 2f = m_1n_2+m_2n_1$$

For this system to have a solution, the determinant of the symmetric matrix of $S$ must vanish:

$$\boxed{\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0}$$

Expanding: $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.

Finding the intersection point

The intersection of the two lines is the solution of:

$$ax + hy + g = 0 \qquad \text{and} \qquad hx + by + f = 0$$

(These are the partial derivatives $\partial S/\partial x = 0$ and $\partial S/\partial y = 0$, halved.)

Angle between the lines

The angle depends only on the homogeneous part $ax^2+2hxy+by^2$:

$$\tan\theta = \frac{2\sqrt{h^2-ab}}{a+b}$$

The terms $2gx+2fy+c$ translate the pair but do not rotate it.