Angle Between Two Lines – Formula, Solved Examples, FAQs

Angle Between Two Lines

The value of the angle between two lines depends on the slopes produced by the intersecting lines. We will calculate the angle between two non-perpendicular and non-parallel lines because the angle between two perpendicular lines will be 90 degrees, and that of parallel lines will be zero degrees. 

Angle Between Two Lines: Formula

If the angle between two intersecting lines having slopes m_1 \text{ and } m_2 is θ, then the formula for the angle θ is given by \tan \theta=\pm\left(m_{2}-m_{1}\right) /\left(1+m_{1} m_{2}\right)

Angle Between Two Straight Lines Derivation

Consider the diagram shown below:

In this diagram, two lines intersect at a point.

Let the slope of these lines be m_{1} \text { and } m_{2} so 

\tan \theta_{1}=m_{1} \text { and } \tan \theta_{2}=m_{2}

We know that the sum of all the angles in a triangle is equal to 180 degrees.

So, in △ABC, 

\theta +{\theta}_1+\text{x} = 180° (Equation No. 1)

Also, \text{x} +{\theta}_2 = 180° (since \text{x and } {\theta}_2 produce a straight line)

Substituting this value in equation no. 1

\theta +{\theta}_1+\text{x}=\text{x}+{\theta}_2

⇒ \theta +{\theta}_1={\theta}_2

Or, \theta=\theta_{2}-\theta_{1}

Therefore, \tan \theta = \tan (\theta_2 -\theta 1)

 ⇒ \tan \theta = (\tan \theta_2 -\tan \theta_1)/(1+\tan\theta_1\tan\theta_2)

Now, substitute the values of \tan \theta_{1} \text { and } \tan \theta_{2}\text{ as }m_1\text{ and }m_2, respectively, 

We have \tan \theta = (m_2 -m_1)/(1+m_1m_2)

Note that in this equation, the value of tan θ will be positive if θ is acute and negative if θ is obtuse.

If the Lines are Perpendicular

If the two lines are perpendicular, the angle between them will be 90 degrees. In this case,

      1/\tan\theta=0

 ⇒ (1+m_{1} m_{2})/(m_{2}-m_{1})=0

 ⇒ (1+m_{1} m_{2})=0
 ⇒ m_{1} m_{2}=-1

If the product of the slopes of lines is -1, it represents that the lines are perpendicular.

If the Lines are Parallel

If two lines are parallel, the angle between them will be zero degrees. In this case,

      \tan\theta=0

 ⇒ \left(m_{2}-m_{1}\right) / 1+m_{1} m_{2}=0
⇒ \left(m_{2}-m_{1}\right)=0
⇒ m_{1}=m_{2}

If the slopes of the two lines are equal, it shows that the lines are parallel.  

Examples 

1. If X (2, -1), Y (5, 3), and Z(-2, 6) are three points, find the angle between the straight lines XY and YZ?

The slope of XY is given by

m_{1}=\left(y_{2}-y_{1}\right) /\left(x_{2}-x_{1}\right)
Here, x_{1}=2, y_{1}=-1, x_{2}=5, y_{2}=3

Substituting the values, we get

\mathrm{m}_{1}=\{3-(-1)\} /(5-2)
\mathrm{m}_{1}=(4 / 3)
\text { Hence, } \mathrm{m}_{1}=4 / 3

Similarly, the slope of YZ is given by

m_{2}=(6-3) /(-2-5)
m_{2}=3 /-7
\text { Hence, } m_{2}=-(3 / 7)

We know that the formula for the angle between two lines

\tan \theta=\left(m_{2}-m_{1}\right) /\left(1-m_{1} m_{2}\right)

We will substitute the values of \mathrm{m}_{1} \text { and } \mathrm{m}_{2}

      tan θ = [{3/(-7) – (4/3)} / {(1+ (-3/7)(4/3)}]

 ⇒ tan θ = [(-37/21) / (-3/7)]

 ⇒ tan θ = (37/9)

Therefore,  \theta=\tan^{-1}(37/9)

2. Find the angle between the two lines 4x – 3y = 8 and 2x + 5y = 4.

We will have to convert the equations of both the lines into a slope-intercept form (y = mx + c) so that we can identify the slope.

Starting with the first equation

     4x – 3y = 8

 ⇒ 3y = 4x – 8

 ⇒ y = (4x – 8)/3

 ⇒ y = (4x/3) – (8/3)

 ⇒ y = (4/3)x – (8/3)

Similarly we will convert 2x + 5y = 4 into slope-intercept form.

     2x + 5y = 4

 ⇒ 5y = -2x + 4

 ⇒ y = (-2x +4)/5

 ⇒ y = (-2x/5) + (4/5)

 ⇒ y = (-2/5)x + (4/5)

Now comparing both the equations with mx + c, we get the values of slopes.

m_1= 4/3 = 1.33 and m_2 = (-2/5) = -0.4

We know that the formula for the angle between two lines

\tan \theta=\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right) /\left(1+\mathrm{m}_{1} \mathrm{~m}_{2}\right)
 ⇒ \tan \theta=\{(1.33-(-0.4)\} /\{1+(1.33) \times(-0.4)\}
 ⇒ \tan \theta=(1.73) /(1-0.532)
 ⇒ \tan \theta=(1.73) /(0.468)
 ⇒ \tan \theta=3.696
\therefore \theta=\tan ^{-1}(3.696)