Quadratic Representation of Pair of Straight Lines

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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 if and only if 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.


Proof

Sufficient Condition

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

\(\ds \LL_1: \ \ \) \(\ds l_1 x + m_1 y + n_1\) \(=\) \(\ds 0\)
\(\ds \LL_2: \ \ \) \(\ds l_2 x + m_2 y + n_2\) \(=\) \(\ds 0\)

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.

$\Box$


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 $\CC$.

$\blacksquare$


Examples

Example: $x^2 - x y - 2 y^2 + 2 x + 5 y = 3$

The equation:

$x^2 - x y - 2 y^2 + 2 x + 5 y = 3$

represents the two straight lines embedded in the Cartesian plane:

\(\ds x + y - 1\) \(=\) \(\ds 0\)
\(\ds x - 2 y + 3\) \(=\) \(\ds 0\)


Example: $x^2 + y^2 - 1 = 0$

The equation:

$x^2 + y^2 - 1 = 0$

does not represent two straight lines embedded in the Cartesian plane.


Sources

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