Intersection of Two Lines

Go back to  'Coordinate-Geometry'

We are given two lines \({L_1}\) and \({L_2}\) , and we are required to find the point of intersection (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 (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:

Angle of intersection between two lines

The equations of the two lines in slope-intercept 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.

Inclination of two intersecting lines

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:

Intersection point of two lines

Download practice questions along with solutions for FREE:
Coordinate Geometry
Grade 9 | Questions Set 1
Coordinate Geometry
Grade 9 | Answers Set 1
Coordinate Geometry
Grade 10 | Questions Set 1
Coordinate Geometry
Grade 10 | Answers Set 1
Download practice questions along with solutions for FREE:
Coordinate Geometry
Grade 9 | Questions Set 1
Coordinate Geometry
Grade 9 | Answers Set 1
Coordinate Geometry
Grade 10 | Questions Set 1
Coordinate Geometry
Grade 10 | Answers Set 1
Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum