Point and Angle of Intersection of Two Straight Lines

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 \(\textbf{Art 5 : } \qquad\boxed{{\text{Point of intersection ; Angle of intersection}}}\)

We are given two lines L1 and L2, and we are required to find the point at which they intersect (if they are non-parallel) 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 \({{\text{L}}_{\text{1}}}\equiv {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\,\,\,\,\,\text{and}\,\,\,{{L}_{2}}\equiv {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\) (written in the most general form). Now, let the point of intersection be (x1,  y1). Thus,

\[\begin{align}  & {{a}_{1}}{{x}_{1}}+{{b}_{1}}{{y}_{1}}+{{c}_{1}}=0 \\  & {{a}_{2}}{{x}_{1}}+{{b}_{2}}{{y}_{1}}+{{c}_{2}}=0 \end{align}\]

This system can be solved to get


From this relation we obtain the point of intersection \(\left( {{x}_{1}},{{y}_{1}} \right)\,\,\text{as}\)

\[\boxed{\,\left( {\frac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}},\,\frac{{{c_1}{a_2} - {c_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}\,} \right)\,}\qquad:\quad \text{Point of intersection}\]

To obtain the angle of intersection between these two lines, consider the figure below:

The equation of the two lines in slope-intercept form are

\[\begin{align}& \,\,\,\,\,\,\,\,\,{{y}_{1}}=\left( -\frac{{{a}_{1}}}{{{b}_{1}}} \right)x+\left( \frac{{{c}_{1}}}{{{b}_{1}}} \right)={{m}_{1}}x+\left( \frac{{{c}_{1}}}{{{b}_{1}}} \right)\,\,\text{where}\,\,{{m}_{1}}=-\frac{{{a}_{1}}}{{{b}_{1}}} \\ & \text{and}\,\,{{y}_{2}}=\left( -\frac{{{a}_{2}}}{{{b}_{2}}} \right)x+\left( \frac{{{c}_{2}}}{{{b}_{2}}} \right)={{m}_{2}}x+\left( \frac{{{c}_{2}}}{{{b}_{2}}} \right)\,\,\text{where}\,\,{{m}_{2}}=-\frac{{{a}_{2}}}{{{b}_{2}}} \end{align}\]

Note in Fig - 15 that \(\theta ={{\theta }_{2}}-{{\theta }_{1}}\) and thus

\[\begin{align} & \tan \theta =\tan \left( {{\theta }_{2}}-{{\theta }_{1}} \right) \\\\  & \qquad =\frac{\tan {{\theta }_{2}}-\tan {{\theta }_{1}}}{1+\tan {{\theta }_{1}}\tan {{\theta }_{2}}} \\\\  & \qquad =\frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}\qquad \qquad \qquad ...\text{ }\left( 1 \right) 

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 (1) above, if the expression \(\begin{align}\frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}\end{align}\) 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

\[\boxed{\theta  = {{\tan }^{ - 1}}\left| {\frac{{{m_2}\,\, - \,{m_1}}}{{1 + {m_1}\,{m_2}}}} \right|\,}\,\,\,\,:\qquad {\mathbf{Acute}}{\text{ }}{\mathbf{angle}}{\text{ }}{\mathbf{between}}{\text{ }}{\mathbf{the}}{\text{ }}{\mathbf{two}}{\text{ }}{\mathbf{lines}}\]

From this relation, we can easily deduce the conditions on m1 and m2 such that the two lines L1 and L2 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, \(\begin{align}\theta = \frac{\pi }{2}\end{align}\) so that \(\cot \theta  = 0;\) this can happen if  \(1 + {m_1}{m_2} = 0\,\quad or\,\quad{m_1}{m_2} =  - 1.\) Thus,

\[\begin{align}& \boxed{{m_1} = {m_2}:{\text{for}}\,{\text{parallel}}\,{\text{lines}}}  \\&\qquad\qquad\qquad and \\&\boxed{{m_1}\,{m_2} =  - 1\,\,:\,\,{\text{for}}\,{\text{perpendicular}}\,{\text{lines}}}\end{align} \]

If the lines L1 and L2 are given in the general form given in the general form  \(ax + by + c = 0,\) the slope of this line is\(\begin{align}m = - \frac{a}{b}\end{align}\) so that the condition for L1 and L2 to be parallel becomes\(\begin{align} - \frac{{{a_1}}}{{{b_1}}} = - \frac{{{a_2}}}{{{b_2}}}\,\,{\text{or}}\,\,{a_1}{b_2} = {a_2}{b_1}\end{align}\) and the condition for Land L2 to be perpendicular becomes \(\begin{align}\frac{{{a_1}{a_2}}}{{{b_1}{b_2}}} = - 1\end{align}\) or \({a_1}{a_2} + {b_1}{b_2} = 0\) .

For example, the line \({L_1} \equiv x + y = 1\) is perpendicular to the line \({L_2} \equiv x - y = 1\) because the slope of L is –1 while the slope of L2 is 1.

As another example, the line \({L_1} \equiv x - 2y + 1 = 0\) is parallel to the line  \({L_2} \equiv x - 2y - 3 = 0\) because the slope of both the lines is \(\begin{align}m = \frac{1}{2}.\end{align}\)

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