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