# Angle Between Two Lines

Let two straight lines *l*_{1} and *l*_{2} be given in the *xy*-plane. Suppose their equations are defined in slope-intercept form:

where *k*_{1} and *k*_{2} are the slopes of the lines, and *b*_{1} and *b*_{2} are their *y*-intercepts, respectively.

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

The second angle is equal to *π − φ*.

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

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

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

we get

So one of the angles between the given lines is

The other angle is

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

Plug in the known coefficients:

So the angle between these lines is

### 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

We obtain an equation for determining *k*:

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:

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

The final answer is

### 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.

Equating the right-hand sides, we get

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

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 \(3.\)

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:

This quadratic equation has the roots:

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 \({\left({-6,-1}\right)}:\)

So, the desired equation of the bisector is

or