In this article, we’ll learn how to determine the equation of the bisector of the angle between two lines whose equations are given. For example, suppose that the equations of lines g and h are y=\frac{7}{2} x and y = \frac{x}{2}, respectively. As Figure 1 shows, they are intersected lines.

Figure 1

In the figure, the green and black lines represent the lines g and h, respectively.

 

An angle bisector is the line or line segment that divides the angle into two equal parts. In Figure 1, the red line is an angle bisector. It divides the angle formed by the green and black lines. What is the equation of the angle bisector? This article addresses the question.

 

Let’s first derive the formula …

Given:

  1. line g with equation A1x + B1y + C1 = 0,
  2. line h with equation A2x + B2y + C2 = 0,
  3. line k, which is a bisector of the angle between g and h

To be sought: the equation of k

Answer

Let P0(x0,y0) be any point that lies on k. Let A be on line g such that \overline{AP_0} \: \bot \: g and B on line h such that \overline{BP_0} \: \bot \: h. Consequently, \vert \overline{AP_0} \vert = distance from P0 to line g and \vert \overline{BP_0} \vert = distance from P0 to line h. (See Figure 2.)

Figure 2

In Figure 2, O is the point of intersection of the lines g and h. Note that \overline{OP}_0 in ΔOAP0 is of an equal length as \overline{OP}_0 in ΔOBP0 and ∠AOP0 = ∠BOP0. As a consequence, ΔOAP0 \cong ΔOBP0. Therefore, \vert \overline{AP_0} \vert = \vert \overline{BP_0} \vert. By applying the formula for the distance between a point and a line, we have the following.

\frac{\vert A_1 x_0 + B_1 y_0 + C_1 \vert}{\sqrt{{A_1}^2 + {B_1}^2}} = \frac{\vert A_2 x_0 + B_2 y_0 + C_2 \vert }{\sqrt{{A_2}^2 + {B_2}^2}}

\frac{A_1 x_0 + B_1 y_0 + C_1}{\sqrt{{A_1}^2 + {B_1}^2}} = \pm \frac{A_2 x_0 + B_2 y_0 + C_2}{\sqrt{{ A_2}^2 + {B_2}^2}}

Since P0(x0,y0) is any point in k, the last equation above holds for all points on the line k. As a result, the equation of k can be expressed as:

\frac{A_1 x + B_1 y + C_1}{\sqrt{{A_1}^2 + {B_1}^2}} = \pm \frac{A_2 x + B_2 y + C_2}{\sqrt{{ A_2}^2 + {B_2}^2}} ……………………………………………………………………………………….. (*)

Note that from (*) it seems that there are actually two possible bisectors.

For every pair of intersected lines there are always two angle bisectors.

The other angle bisector is represented by the yellow line in Figure 3.

Figure 3

 

It was said earlier that there were two angle bisectors for each pair of intersected lines. Furthermore, it can be proved that the angle bisectors are perpendicular to each other.

Now we are ready to solve the problem at the beginning of this article.

Given:

  1. line g with equation y=\frac{7}{2} x,
  2. line h with equation y = \frac{x}{2},
  3. line k, which is a bisector of the angle between g and h

To be sought: the equation of k

Answer

Note that the lines g and h can be expressed in the form of Ax + By + C = 0 as follows.

g ≡ 7x – 2y = 0

h ≡   x – 2y = 0

Substituting A1 = 7, B1 = -2, C1 = 0, A2 = 1, B2 = -2, and C2 = 0 into (*), we get:

\frac{7x-2y}{\sqrt{7^2+(-2)^2}} = \pm \frac{x-2y}{\sqrt{1^2+(-2)^2 }}

There are two angle bisectors. Their equations are determined as follows.

 

First angle bisector: (See Figure 4)

\frac{7x-2y}{\sqrt{7^2+(-2)^2}} = \frac{x-2y}{\sqrt{1^2+(-2)^2}}

\Leftrightarrow y = \frac{3 - \sqrt{265}}{16} x

Figure 4

Second angle bisector: (See Figure 5)

\frac{7x-2y}{\sqrt{7^2+(-2)^2}} =  -  \frac{x-2y}{\sqrt{1^2+(-2)^2} }

\Leftrightarrow y  =  \frac{3 + \sqrt{265}}{16} x

Figure 5

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