Intersection of Two Lines

Intersection of Two Lines
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This page is all about the intersection of two lines.

In this mini-lesson, we will learn in detail about, How to find the point of intersection of two lines?

We all are familiar with two-dimensional coordinate geometry from the earlier classes.

Primarily, it is a combination of algebra and geometry.

The renowned French philosopher and mathematician, Rene Descartes, published a book, La Géométry in 1637. In this book, he introduced the world with the systematic study of geometry from the use of algebra.

"The intersection of two lines" is an introductory topic under two-dimensional coordinate geometry.

Come, let us learn in detail about how to find the point of intersection of two lines.

In this lesson, we will also learn about intersecting lines examples, the point of intersection formula, and the intersection point calculator. 

Lesson Plan

What Does Intersection of Two Lines Mean?

Intersecting Lines Definition

When two lines share exactly one common point, they are called intersecting lines.

The intersecting lines share a common point. And, this common point that exists on all intersecting lines is called the point of intersection.

Here, lines \(A\) and \(B\) intersect at point \(O\), which is the point of intersection.

Point of intersection of lines A and B

Examples In Real Life

Common examples of intersecting lines in real life include a pair of scissors, a folding chair, a road cross, signboard, etc. 

Intersection Lines in real life - a pair of scissors, a folding chair, a road cross, signboard

Let's consider the following case.

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

Evaluating the point of intersection involves solving two simultaneous linear equations.

Let the equations of the two lines be (written in the general form):

\[\begin{array}{l}{a_1}x + {b_1}y + {c_1} = 0\\{a_2}x + {b_2}y + {c_2} = 0\end{array}\]

Point of Intersection and Angle of Intersection - graph

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 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 can obtain the point of intersection \(\left( {{x_0},{y_0}} \right)\) as

\[\left( {{x_0},{y_0}} \right) = \left( {\frac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}},\frac{{{c_1}{a_2} - {c_2}{a_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}} \right)\]

Intersection Point Calculator

To obtain the angle of intersection between these two lines, consider the figure shown above.

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.

Conditions for Two Lines to be Parallel or Perpendicular 

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

For the two lines to be perpendicular lines, \(\theta  = \frac{\pi }{2}\), 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 in the general form \(ax + by + c = 0\), the slope of this line is \(m =  - \frac{a}{b}\).

Condition for Two Lines to be Parallel

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 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 \(m = \frac{1}{2}\).

Condition for Two Lines to be Perpendicular

The condition for \({L_1}\) and \({L_2}\) to be perpendicular is:

\[\begin{align}&{m_1}{m_2} = - 1\, \Rightarrow \,\left( { - \frac{{{a_1}}}{{{b_1}}}} \right)\left( { - \frac{{{a_2}}}{{{b_2}}}} \right) = - 1\,\\ &\qquad\qquad\;\;\;\; \Rightarrow \,\,{a_1}{a_2} + {b_1}{b_2} = 0\end{align}\]


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.

Properties of Intersecting Lines

  • The intersecting lines (two or more) always meet at a single point.
  • The intersecting lines can cross each other at any angle. This angle formed is always greater than \(0^{\circ}\) and less than \(180^{\circ}\). 
  • Two intersecting lines form a pair of vertical angles. The vertical angles are opposite angles with a common vertex  (which is the point of intersection). 

Vertical angles of line L1 and L2

Here,\(\angle a\) and \(\angle c\) are vertical angles and are equal. 

Also, \(\angle b\) and \(\angle d\) are vertical angles and equal to each other. 

\(\angle a+\angle d\) = straight angle = \(180^{\circ}\)

Challenge your math skills

Challenging Questions

  1. Reduce the following equations into slope-intercept form
    (i) \(x+7y=0\)                              (ii) \(3x+2y-12\)
  2. If the angle between two lines is \(\frac{\pi}{4}\) and slope of one the lines is \(\frac{1}{2}\), find the slope of the other line.

Solved Examples

Example 1



 Find the point of intersection and the angle of intersection for the following two lines:

Point of intersection - graph

\[\begin{array}{l}x - 2y + 3 = 0\\3x - 4y + 5 = 0\end{array}\]


We use Cramer’s rule to find the point of intersection:

\[\begin{align}&\frac{x}{{ - 10 - \left( { - 12} \right)}} = \frac{-y}{{5 -9 }} = \frac{1}{{ - 4 - \left( { - 6} \right)}}\\&\Rightarrow \,\,\,\frac{x}{2} = \frac{y}{4} = \frac{1}{2}\\&\Rightarrow \,\,\,x = 1,\,\,\,y = 2\end{align}\]
Also, you can check your answer with the help of cuemath's Intersection Point Calculator

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: 

\[\begin{align}&\tan \theta  = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right| = \left| {\frac{{\frac{3}{4} - \frac{1}{2}}}{{1 + \frac{3}{8}}}} \right| = \frac{2}{{11}}\\&\Rightarrow \,\,\,\theta  = {\tan ^{ - 1}}\left( {\frac{2}{{11}}} \right) \approx {10.3^\circ}\end{align}\]

\(\therefore\) Point of intersection is \((1,2)\).
Angle of intersection is \(\theta=\tan ^{ - 1}\left( {\frac{2}{11}} \right)\).

Example 2



 Find the equation of a line perpendicular to the line \(x - 2y + 3 = 0\) and passing through the point\((1, \;-2)\)

Point of intersection - graph


Given line  \(x - 2y + 3 = 0\) can be written as

\[{y} = \frac{1}{2}x\,\,\, + \frac{3}{2}\]

Slope of the line \((1)\) is \({m_1}= \frac{1}{2}\).

Therefore, slope of the line perpendicular to line \((1)\) is

\[{m_2} = -\frac{1}{{m_1}},\,\,\, = -2\]

Equation of a line perpendicular to the line \(x - 2y + 3 = 0\) and passing through the point\((1, \;-2)\) is

\[y-(-2) = -2(x-1),\,\,\, \,\,\, \text{or} \,\,\,y= -2x\]

Which is the required equation

\(\therefore\) Equation of the required line is \(y= -2x\)

important notes to remember
Important Notes
  • An acute angle \(\theta \) between lines \(L_1\) and \(L_2\) with slopes \(m_1\) and \(m_2\) is given by

    \[\theta  = {\tan ^{ - 1}}\left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\]

  • If the lines \({L_1}\) and \({L_2}\) are given in the general form \(ax + by + c = 0\), the slope of this line is \(m =  - \frac{a}{b}\) .

    The condition for two lines \({L_1}\) and \({L_2}\) to be parallel is:

    \({m_1} = {m_2}\)

    The condition for two lines \({L_1}\) and \({L_2}\) to be perpendicular is:

    \({m_1}{m_2} =  - 1\)

Interactive Questions

Here are a few activities for you to practice.

Select/Type your answer and click the "Check Answer" button to see the result.


Let's Summarize

The mini-lesson targeted the fascinating concept of "The intersection of two lines." The math journey around "The intersection of two lines" starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

About Cuemath

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

Frequently Asked Questions (FAQs)

1. How do I find the point of intersection of two lines?

Here's the summary of our methods:

Get the two equations for the lines into slope-intercept form. That is, have them in this form: y = mx + b.
Set the two equations for y equal to each other.
Solve for x. This will be the x-coordinate for the point of intersection.
Use this x-coordinate and substitute it into either of the original equations for the lines and solve for y. This will be the y-coordinate of the point of intersection.
To verify, substitute the x-coordinate into the other equation and you should get the same y-coordinate.
You now have the x-coordinate and y-coordinate for the point of intersection.

2. What does the intersection of two lines represent?

When the lines intersect, the point of intersection is the only point that the two graphs have in common, so the coordinates of that point are the solution for the two variables used in the equations. When the lines are parallel, there are no solutions.

3. What is the condition for intersection of two lines?

A necessary condition for two lines to intersect is that they are in the same plane i.e., they are not skew lines.

4. Can two planes intersect in a line?

They cannot intersect at only one point because planes are infinite. Furthermore, they cannot intersect over more than one line because planes are flat. One way to think about planes is to try to use sheets of paper and observe that the intersection of two sheets would only happen at one line.

5. How many solutions do same lines have?

A system of linear equations usually has a single solution, but sometimes it can have no solution (parallel lines) or infinite solutions (same line).

6. When two lines intersect how many angles are formed?

When two lines intersect, four angles are formed.

7. Do parallel lines have a solution?

Since parallel lines never cross, there can be no intersection; that is, for a system of equations that graphs as parallel lines, there can be no solution. This is called an "inconsistent" system of equations, and it has no solution.

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