Angle Between Two Lines

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Let two straight lines l₁ and l₂ be given in the xy-plane. Suppose their equations are defined in slope-intercept form:

\[\ell_1:\;y = k_1x + b_1,\;\;\;\;\ell_2:\;y = k_2x + b_2,\]

where k₁ and k₂ are the slopes of the lines, and b₁ and b₂ are their y-intercepts, respectively.

One of the angles φ between these lines is determined by the formula

\[\tan\varphi = \frac{k_2 - k_1}{1+k_1k_2}\]

The second angle is equal to π − φ.

Angle between two straight lines — Figure 1.

Note that the angle between two lines is defined as the smaller angle between the two angles formed by the lines. If angle φ is acute then the angle π − φ is obtuse and vice versa.

Now let the lines be defined by equations in general form as

\[\ell_1:\;a_1x + b_1y + c_1 = 0,\;\;\;\;\ell_2:\;a_2x + b_2y + c_2 = 0,\]

In this case, the tangent of the angle φ between the straight lines is given by the formula

\[\tan\varphi = \frac{a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2}\]

The examples below help in understanding the concept of the angle between two lines.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find the angle between the lines \({y = 2x - 1}\) and \({y = \frac{1}{2}x + 1}.\)

Example 2

Find the angle between two lines \({5x - y + 2 = 0}\) and \({2x - 3y + 4 = 0}.\)

Example 3

Write equations of the lines passing through the origin and inclined at an angle of \(45^\circ\) to the line \({y = 2x + 1}.\)

Example 4

Find the equation of the bisector of the acute angle between the lines \(x - 7y - 1 = 0\) and \(x + y + 7 = 0.\)

Example 1.

Find the angle between the lines \({y = 2x - 1}\) and \({y = \frac{1}{2}x + 1}.\)

Solution.

The given lines have the slopes \(k_1 = 2\) and \(k_2 = \frac{1}{2}.\) Using the formula

\[\tan\varphi = \frac{k_2 - k_1}{1+k_1k_2},\]

we get

\[\tan\varphi = \frac{\frac{1}{2} - 2}{1+\frac{1}{2}\cdot2} = \frac{-\frac{3}{2}}{1+1} = -\frac{3}{4}.\]

So one of the angles between the given lines is

\[\varphi = \arctan\left({-\frac{3}{4}}\right) = -\arctan\frac{3}{4}.\]

The other angle is

\[\pi - \varphi = \pi - \left({-\arctan\frac{3}{4}}\right) = \pi + \arctan\frac{3}{4} = \arctan\frac{3}{4} \approx 36.9^\circ\]

Hence, the angle between the lines is \(\arctan\frac{3}{4}.\)

Example 2.

Find the angle between two lines \({5x - y + 2 = 0}\) and \({2x - 3y + 4 = 0}.\)

Solution.

These lines are represented by equations in general form. Therefore, we use the formula

\[\tan\varphi = \frac{a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2}.\]

Plug in the known coefficients:

\[\tan\varphi = \frac{2\cdot\left(-1\right)- 5\cdot\left(-3\right)}{5\cdot2 + \left(-1\right)\cdot\left(-3\right)} = \frac{-2+15}{10+3} = \frac{\cancel{13}}{\cancel{13}} = 1.\]

So the angle between these lines is

\[\varphi = \arctan1 = 45^\circ.\]

Example 3.

Write equations of the lines passing through the origin and inclined at an angle of \(45^\circ\) to the line \({y = 2x + 1}.\)

Solution.

Let the equation of the required straight line have the form \(y = kx + b.\) Since the angle between the lines is known, we can write

\[\tan\varphi = \tan 45^\circ = 1 = \frac{k - 2}{1 + 2k\cdot2} = \frac{k-2}{1+2k}.\]

We obtain an equation for determining k:

\[1 = \frac{k-2}{1+2k},\;\Rightarrow k - 2 = 1 + 2k, \;\Rightarrow k_1 = -3.\]

So, the slope of the first line is equal to \(-3\). Since the line passes through the origin, its y-intercept is zero, that is the equation of the line is \(y = -3x.\)

To find the equation of the second line, let's make the above expression equal to π − φ. This yields:

\[\frac{k-2}{1+2k} = \tan\left({\pi-\varphi}\right) = \tan 135^\circ = -1, \;\Rightarrow k-2 = -\left({1+2k}\right),\;\Rightarrow 3k = 1, \;\Rightarrow k_2 = \frac{1}{3}.\]

Hence, the equation of the second line is \(y = \frac{1}{3}x.\)

Two lines inclined to the line y=2x+1 at an angle of 45 degrees — Figure 2.

The final answer is

\[y = -3x,\;y = \frac{1}{3}x.\]

Example 4.

Find the equation of the bisector of the acute angle between the lines \(x - 7y - 1 = 0\) and \(x + y + 7 = 0.\)

Solution.

First we rewrite the equations of our lines in slope-intercept form and find the point of their intersection.

\[\begin{cases} x - 7y - 1 = 0 \\ x + y + 7 = 0 \end{cases}, \;\Rightarrow \begin{cases} y = \frac{1}{7}x - \frac{1}{7} \\ y = -x - 7 \end{cases}.\]

Equating the right-hand sides, we get

\[\frac{1}{7}x - \frac{1}{7} = -x - 7, \;\Rightarrow x - 1 = -7x - 49, \;\Rightarrow 8x = -48, \;\Rightarrow x = -6.\]

Hence, the \(y-\)coordinate of the intersection point is

\[y = - x - 7 = -\left({-6}\right) - 7 = 6 - 7 = -1.\]

We see that the lines intersect at the point \({\left({-6,-1}\right)}.\) Their slope coefficients are, respectively, \(k_1 = \frac{1}{7}\) and \(k_2 = -1,\) so the straight lines look like shown in Figure \(3.\)

Finding the angle bisector equation — Figure 3.

Let the bisector equation have the form \(y = kx + b.\) To find the slope \(k\) of the bisector we equate the tangents of the angles \(\varphi\) between the second line and the bisector and between the bisector and the first line:

\[\tan\varphi = \frac{k_2 - k}{1 + k_2k} = \frac{k - k_1}{1 + kk_1},\;\Rightarrow \frac{-1 - k}{1 - k} = \frac{k - \frac{1}{7}}{1 + \frac{1}{7}k},\;\Rightarrow \frac{k + 1}{k - 1} = \frac{7k - 1}{7 + k},\]

\[\Rightarrow \left({k+1}\right)\left({7+k}\right) = \left({7k-1}\right)\left({k-1}\right)\;\Rightarrow 7k+7+k^2+k = 7k^2-k-7k+1,\;\Rightarrow 6k^2 - 16k - 6 = 0, \;\Rightarrow 3k^2 - 8k - 3 = 0.\]

This quadratic equation has the roots:

\[3k^2 - 8k - 3 = 0,\;\Rightarrow D = \left({-8}\right)^2 - 4\cdot3\cdot\left({-3}\right) = 100, \;\Rightarrow k = \frac{8 \pm \sqrt{100}}{6} = 3,-\frac{1}{3}.\]

We will choose the value \({-\frac{1}{3}}\) since the bisector divides an acute angle. To find the value of \(b,\) we plug in the coordinates of the point of intersection \({\left({-6,-1}\right)}:\)

\[y = -\frac{1}{3}x + b, \;\Rightarrow -1 = -\frac{1}{3}\cdot\left({-6}\right) + b,\;\Rightarrow b = -3.\]

So, the desired equation of the bisector is

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

\[x + 3y + 9 = 0.\]