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I have three points,$$A =[A_x,A_y]\,,\, B =[B_x,B_y]\,,\,C =[C_x,C_y]$$

How could one calculate the equation for the line that bisects the angle $\,\angle ABC\,$ (eg., passing through $B$)?

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It'd be nice to have some insights, ideas, self-work from you. For example, can you characterize the bisector of an angle as the locus of some points that...? –  DonAntonio Nov 16 '12 at 19:54
    
I'm not quite sure how else to describe an angle bisector other than that it cuts the angle into two equal parts. As far as stuff I have tried... I thought maybe the midpoint of AC would always be on the angle bisector (not so), and i also thought maybe I could find the length of say AB, then find the point D on the line AC such that the length of AD = AB, then find the midpoint between the segment BD –  AndyPerlitch Nov 16 '12 at 20:01
    
Read here ditutor.com/line/equation_bisector.html what I meant in my comment and in my answer below. –  DonAntonio Nov 16 '12 at 20:04

2 Answers 2

up vote 1 down vote accepted

Start with a translation -- i.e., shift things to a slightly modified coordinate system so that $B$ is the origin:

$$ A \mapsto \left[A_x-B_x, A_y-B_y\right] = A'\\ B \mapsto \left[B_x-B_x, B_y-B_y\right] = B'\\ C \mapsto \left[C_x-B_x, C_y-B_y\right] = C'$$

Compute the angles formed by $A'$ and $C'$ with respect to the positive $x$ axis:

$$\theta_A = \tan^{-1} \frac{A_y-B_y}{A_x-B_x}$$ $$\theta_C = \tan^{-1} \frac{C_y-B_y}{C_x-B_x}$$

Compute the angle between $A'$ and $C'$:

$$\theta = \cos^{-1} \left( \frac{A'\cdot B'}{\|A'\|\|B'\|} \right)$$

Divide by two, and add this angle to the smaller of $\theta_A$ and $\theta_C$:

$$\phi = \min \left\{ \theta_A, \theta_C\right\}+\frac{\theta}{2}$$

Find a unit vector that has this angle in the shifted coordinate system, i.e.

$$\begin{align*} \tan \phi &= \frac{D_y}{D_x} \\ D_x^2+D_y^2 &= 1 \end{align*}$$

Alternatively, ignore the fact that it must be a unit vector, and just set $D_x = 1$ and compute $D_y$--it will still be on the same line, no matter what.

Finally, shift things back to original coordinates:

$$D = \left[ D_x + B_x, D_y + B_y \right].$$


Example:

$$A = [3,3] \\ B = [2,2] \\ C = [1,3]$$

Then,

$$A' = [3-2,3-2] = [1,1] \\ C' = [1-2,3-2] = [-1,1],$$ $$ \theta = \cos^{-1} \frac{ 1\cdot (-1)+1\cdot 1}{\sqrt{1^2+1^2}\sqrt{(-1)^2+1^2}} = \cos^{-1} 0 = \pi/2\ \;\;\; (\text{90 degrees}),$$ $$ \theta_A = \tan^{-1} 1 = \pi/4\ \;\;\; (\text{45 degrees})$$ $$ \theta_C = \tan^{-1} -1 = 3\pi/4\ \;\;\; (\text{135 degrees})$$

The smaller of $\theta_A$, $\theta_C$ is $\theta_A$, so finally,

$\phi = \frac{\pi}{4}+\frac{1}{2}\theta = \frac{\pi}{4}+\frac{\pi}{4} = \frac{\pi}{2}$

as expected.

If $\phi = \pi/2$, then this is a vector pointing along the y-axis, so let $D' = (0,1)$, which leads to $D = D'+B = (0+2,1+2) = (2,3).$

Plot $A, B, C, and D$ and you will see that this is the desired result.

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This probably isn't the easiest or fastest way, but I tried to write it in a way that makes the geometrical process clear. –  Arkamis Nov 16 '12 at 20:05
    
Ok i think i get it... So D would be a point on the angle bisector, then just compute the line that passes through B and D? –  AndyPerlitch Nov 16 '12 at 20:11
    
Yep, exactly. Note that $A' \cdot C' = ((A_x-B_x)(C_x-B_x) + (A_y-B_y)(C_y-B_y))$ denotes the dot product between the vectors. –  Arkamis Nov 16 '12 at 20:15
    
ok and does ||A'|| ||B'|| = (|AxBx| + |AyBy|)? apologies, i am not familiar with LaTeX formatting... –  AndyPerlitch Nov 16 '12 at 20:17
    
No, that means the vector norm: $\|A'\| = \sqrt{(A_x-B_x)^2+(A_y-B_y)^2}$ –  Arkamis Nov 16 '12 at 20:28

What about setting the equations for a general point $\,(x,y)\,$ in the plane being at the same distance from line $\,AB\,$ and from line $\,BC\,$?

Why would this suffice and, thus, work?

Added: We have

$$(1)\,\,\text{The line}\,\,AB:\,m_{AB}=\frac{B_y-A_y}{B_x-A_x}\Longrightarrow m_{AB}x-y-m_{AB}B_x+B_y=0$$

$$(2)\,\,\text{The line}\,\,CB:\,m_{CB}=\frac{B_y-C_y}{B_x-C_x}\Longrightarrow y=m_{CB}(x-B_x)+B_y$$

Now, if $\,(x,y)\,$ is a point equidistant from the above two line, then we have that

$$(3)\;\;\;\;\frac{|m_{AB}x-y-m_{AB}B_x+B_y|}{\sqrt{m_{AB}^2+1}}=\frac{|m_{CB}x-y-m_{CB}B_x+B_y|}{\sqrt{m_{CB}^2+1}}$$

Solving the above gives you the bisector's line equation.

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