Intersection of Two Lines
We are given two lines \({L_1}\) and \({L_2}\) , and we are required to find the point of intersection (if they are nonparallel) and the angle at which they are inclined to one another, i.e., the angle of intersection. Evaluating the point of intersection is a simple matter of solving two simultaneous linear equations. Let the equations of the two lines be (written in the general form):
\[\begin{array}{l}{a_1}x + {b_1}y + {c_1} = 0\\{a_2}x + {b_2}y + {c_2} = 0\end{array}\]
Now, let the point of intersection be \(\left( {{x_0},{y_0}} \right)\). Thus,
\[\begin{array}{l}{a_1}{x_0} + {b_1}{y_0} + {c_1} = 0\\{a_2}{x_0} + {b_2}{y_0} + {c_2} = 0\end{array}\]
This system can be solved using the Cramer’s rule to get
\[\frac{{{x_0}}}{{{b_1}{c_2}  {b_2}{c_1}}} = \frac{{  {y_0}}}{{{a_1}{c_2}  {a_2}{c_1}}} = \frac{1}{{{a_1}{b_2}  {a_2}{b_1}}}\]
From this relation we obtain the point of intersection \(\left( {{x_0},{y_0}} \right)\) as
\[\left( {{x_0},{y_0}} \right) = \left( {\frac{{{b_1}{c_2}  {b_2}{c_1}}}{{{a_1}{b_2}  {a_2}{b_1}}},\frac{{{c_1}{a_2}  {c_2}{a_1}}}{{{a_1}{b_2}  {a_2}{b_1}}}} \right)\]
To obtain the angle of intersection between these two lines, consider the figure below:
The equations of the two lines in slopeintercept form are:
\[\begin{align}&y = \left( {  \frac{{{a_1}}}{{{b_1}}}} \right)x + \left( {\frac{{{c_1}}}{{{b_1}}}} \right) = {m_1}x + {C_1}\\&y = \left( {  \frac{{{a_2}}}{{{b_2}}}} \right)x + \left( {\frac{{{c_2}}}{{{b_2}}}} \right) = {m_2}x + {C_2}\end{align}\]
Note in the figure above that \(\theta = {\theta _2}  {\theta _1}\), and thus
\[\begin{align}&\tan \theta = \tan \left( {{\theta _2}  {\theta _1}} \right) = \frac{{\tan {\theta _2}  \tan {\theta _1}}}{{1 + \tan {\theta _1}\tan {\theta _2}}}\\&\qquad\qquad\qquad\qquad\;\;= \frac{{{m_2}  {m_1}}}{{1 + {m_1}{m_2}}}\end{align}\]
Conventionally, we would be interested only in the acute angle between the two lines and thus we have to have \(\tan \theta \) as a positive quantity. So in the expression above, if the expression \(\frac{{{m_2}  {m_1}}}{{1 + {m_1}{m_2}}}\) turns out to be negative, this would be the tangent of the obtuse angle between the two lines; thus, to get the acute angle between the two lines, we use the magnitude of this expression.
Therefore, the acute angle \(\theta \) between the two lines is
\[\theta = {\tan ^{  1}}\left {\frac{{{m_2}  {m_1}}}{{1 + {m_1}{m_2}}}} \right\]
From this relation, we can easily deduce the conditions on \({m_1}\) and \({m_2}\) such that the two lines \({L_1}\) and \({L_2}\) are parallel or perpendicular.

If the lines are parallel, \(\theta = 0\) , so that \({m_1} = {m_2}\) , which is intuitively obvious since parallel lines must have the same slope.

For the two lines to be perpendicular, \(\theta = \frac{\pi }{2}\), so that \(\cot \theta = 0\); this can happen if \(1 + {m_1}{m_2} = 0\) or \({m_1}{m_2} =  1\) .
If the lines \({L_1}\) and \({L_2}\) are given in the general form given in the general form \(ax + by + c = 0\), the slope of this line is \(m =  \frac{a}{b}\) . Thus, the condition for \({L_1}\) and \({L_2}\) to be parallel is:
\[{m_1} = {m_2}\,\,\, \Rightarrow \,\,\,  \frac{{{a_1}}}{{{b_1}}} =  \frac{{{a_2}}}{{{b_2}}}\,\,\, \Rightarrow \,\,\,\frac{{{a_1}}}{{{b_1}}} = \frac{{{a_2}}}{{{b_2}}}\]
The condition for \({L_1}\) and \({L_2}\) to be perpendicular is:
\[\begin{align}&{m_1}{m_2} =  1\,\,\, \Rightarrow \,\,\,\left( {  \frac{{{a_1}}}{{{b_1}}}} \right)\left( {  \frac{{{a_2}}}{{{b_2}}}} \right) =  1\,\\ &\qquad\qquad\;\;\;\;\;\; \Rightarrow \,\,\,{a_1}{a_2} + {b_1}{b_2} = 0\end{align}\]
For example, the line \({L_1}:x + y = 1\) is perpendicular to the line \({L_2}:x  y = 1\) because the slope of \({L_1}\) is \(  1\) while the slope of \({L_2}\) is 1.
As another example, the line \({L_1}:x  2y + 1 = 0\) is parallel to the line \({L_2}:x  2y  3 = 0\) because the slope of both the lines is \(m = \frac{1}{2}\).
Example 1: Find the point of intersection and the angle of intersection for the following two lines:
\[\begin{array}{l}x  2y + 3 = 0\\3x  4y + 5 = 0\end{array}\]
Solution: We use Cramer’s rule to find out the point of intersection:
\[\begin{align}&\frac{x}{{  10  \left( {  12} \right)}} = \frac{y}{{9  5}} = \frac{1}{{  4  \left( {  6} \right)}}\\&\Rightarrow \,\,\,\frac{x}{2} = \frac{y}{4} = \frac{1}{2}\\&\Rightarrow \,\,\,x = 1,\,\,\,y = 2\end{align}\]
Now, the slopes of the two lines are:
\[{m_1} = \frac{1}{2},\,\,\,{m_2} = \frac{3}{4}\]
If \(\theta \) is the acute angle of intersection between the two lines, we have:
\[\begin{align}&\tan \theta = \left {\frac{{{m_1}  {m_2}}}{{1 + {m_1}{m_2}}}} \right = \left {\frac{{\frac{1}{2}  \frac{3}{4}}}{{1 + \frac{3}{8}}}} \right = \frac{2}{{11}}\\&\Rightarrow \,\,\,\theta = {\tan ^{  1}}\left( {\frac{2}{{11}}} \right) \approx {10.3^\circ}\end{align}\]
The answers can be verified as correct from the following figure: