Quadratic Representation of Pair of Straight Lines

Theorem
Consider the general quadratic equation in $2$ variables:


 * $(1): \quad a x^2 + b x y + c y^2 + d x + e y + f = 0$

Then $(1)$ is the locus of $2$ straight lines in the Cartesian plane it can be expressed in the form:


 * $\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$

where $l_1$, $m_1$, $n_1$, $l_2$, $m_2$ and $n_2$ are real numbers.

Sufficient Condition
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in general form as:

Let $\tuple {x, y}$ be a point on either $\LL_1$ or $\LL_2$.

Then because either $l_1 x + m_1 y + n_1 = 0$ or $l_2 x + m_2 y + n_2 = 0$, it is certainly the case that:
 * $\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$

If $\tuple {x, y}$ is not on either $\LL_1$ or $\LL_2$, then neither $l_1 x + m_1 y + n_1 = 0$ nor $l_2 x + m_2 y + n_2 = 0$ hold.

Hence $l_1 x + m_1 y + n_1 \ne 0$ and $l_2 x + m_2 y + n_2 \ne 0$, and so:
 * $\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} \ne 0$

Hence two straight lines embedded in $\CC$ can be expressed by an equation in the form:
 * $\paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$

as required.

Necessary Condition
Let it be possible to express $(1)$ in the form:


 * $(2): \quad \paren {l_1 x + m_1 y + n_1} \paren {l_2 x + m_2 y + n_2} = 0$

Let $\tuple {x, y}$ satisfy $(2)$.

Then either $l_1 x + m_1 y + n_1 = 0$ or $l_2 x + m_2 y + n_2 = 0$ or both.

Thus $\tuple {x, y}$ is either:
 * on the straight line defined by the equation $l_1 x + m_1 y + n_1 = 0$

or:
 * on the straight line defined by the equation $l_2 x + m_2 y + n_2 = 0$

It follows that $(2)$, and hence $(1)$, is an equation for two straight lines in $\CC$.