# General Equation of a Pair of Lines

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## GENERAL EQUATION OF A PAIR OF LINES

Consider the equations of two arbitrary lines $${L_1}{\text{ and }}{L_2}:$$

\begin{align} {L_1}:{l_1}x + {m_1}y + {n_1} = 0\\ {L_2}:{l_2}x + {m_2}y + {n_2} = 0 \end{align}

The joint equation of the two lines is

\begin{align} &\qquad \qquad \qquad \qquad {L_1}{L_2} = 0\\ \Rightarrow \qquad & \qquad ({l_1}x + {m_1}y + {n_1})({l_2}x + {m_2}y + {n_2}) = 0\\ \Rightarrow \qquad & \quad ({l_1}{l_2}){x^2} + ({l_1}{m_2} + {m_1}{l_2})xy + ({m_1}{m_2}){y^2} + ({n_1}{l_2} + {n_2}{l_1})x + ({n_1}{m_2} + {n_2}{m_1})y + {n_1}{n_2} = 0 \qquad \qquad \qquad \qquad \qquad ...(1) \end{align}

(1) suggests that the most general equation to a pair of straight lines has the form

$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 \qquad \qquad \qquad \qquad ...(2)$

It might be apparent to you that (2) will not always represent a pair of straight lines. For (2) to indeed represent a pair of straight lines, it must be able to be factorised into two linear factors; as an exercise for the reader, show that (2) can be expressed as a product of linear factors if the following condition is satisfied:

\begin{align} & abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0\\ \Rightarrow \qquad & \left| {\begin{array}{*{20}{c}} a&h&g\\ h&b&f\\ g&f&c \end{array}} \right| = 0 \end{align}

We now re-evaluate the conditions for parallel and perpendicular lines, in the general case :

Let $${L_1}{\text{ and }}{L_2}$$ be two lines with slopes $${s_1}{\text{ and }}{s_2};$$ their equations have already been mentioned above. $${L_1}{\text{ and }}{L_2}$$ are parallel if

\begin{align} & {s_1} = {s_2}\\ \Rightarrow \qquad & \frac{{ - {l_1}}}{{{m_1}}} = \frac{{ - {l_2}}}{{{m_2}}}\\ \Rightarrow \qquad & {l_1}{m_2} = {l_2}{m_1}\\ \Rightarrow \qquad & {\left( {{l_1}{m_2} - {l_2}{m_1}} \right)^2} = 0\\ \Rightarrow \qquad & {\left( {{l_1}{m_2} + {l_2}{m_1}} \right)^2} = 4\,{l_1}\,{l_2}\,{m_1}\,{m_2} \qquad\qquad\qquad\qquad ...(3) \end{align}

From (2), observe that the constraint in (3) can be specified as

\begin{align} \qquad & 4{h^2} = 4ab \hfill \\ \Rightarrow \qquad & \boxed{{h^2} = ab} \qquad : \qquad \operatorname{Parallel} \;lines \hfill \\ \end{align}

This is the same condition as the one for the homogenous case.

For $${L_1}{\text{ and }}{L_2}$$ to be perpendicular,

\begin{align} & {s_1} \cdot {s_2} = - 1 \hfill \\ \Rightarrow \qquad & \frac{{- {l_1}}}{{{m_1}}} \cdot \frac{{ - {l_2}}}{{{m_2}}} = - 1 \hfill \\ \Rightarrow \qquad & {l_1}{l_2} + {m_1}{m_2} = 0 \hfill \\ \Rightarrow \qquad & \boxed{a + b = 0} \qquad : \qquad \text{Perpendicular lines} \hfill \\ \end{align}

Again, this condition is the same as the one in the homogenous case.

If fact, you can verify that the angle $$\theta$$ subtended between the two lines is also given by the same formula as in the homogenous case, i.e.,

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

That these formulae in the homogenous and the general case are the same should be obvious since the slope of any line is independent of the constant term appearing in its equation.