# Help with equation of the bisector of an angle in a triangle

• Aug 23rd 2009, 07:37 AM
ninni
Help with equation of the bisector of an angle in a triangle
Find the equation of the bisector of angle A of the triangle, the coordinates of whose vertices are A(4,3) , B(0,0) and C(2,3). Please give its solution. Thanks a lot for your help.
• Aug 23rd 2009, 08:15 AM
red_dog
The equation of the side AB: $\displaystyle 3x-4y=0$

The equation of the side AC: $\displaystyle y-3=0$

Let M(x,y) be a point of the bisector. Then $\displaystyle d(M,AB)=d(M,AC)$

$\displaystyle \frac{|3x-4y|}{5}=|y-3|$

We get two equations: $\displaystyle 2x+4y-15=0$ and $\displaystyle 8x-4y-15=0$

These are the equations of both bisectors, interior and exterior. To find which of them is the interior bisector, plug the coordinates of B and C in the left side member of the equations and we have to get two values of opposite signs.
• Aug 23rd 2009, 08:24 AM
Plato
Here is a hint.
If $\displaystyle \overrightarrow a \;\& \,\overrightarrow b$ are vectors then the bisector of the angle between them is $\displaystyle \left\| {\overrightarrow b } \right\|\overrightarrow a + \left\|{\overrightarrow a } \right\|\overrightarrow b$.
• Aug 23rd 2009, 08:27 AM
Soroban
Hello, ninni!

I assume you made a sketch . . .
We need some geometry and a half-angle identity.

Quote:

Find the equation of the bisector of angle $\displaystyle A$ of $\displaystyle \Delta ABC.$
The coordinates of the vertices are: .$\displaystyle A(4,3),\; B(0,0),\;C(2,3)$

Code:

|    C                  A
|(2,3)*    *    *    * (4,3)
|    *          o  *
|  *  o      *
|  *        *          o D
| *    *      o
|*  *  o
- - * - - - - - - - - - - - - - - X
B

Let $\displaystyle BD$ = bisector of $\displaystyle \angle ABX.$

Note that $\displaystyle \angle A \:=\:\angle CAB \:=\:\angle ABX$
Hence, the bisector of $\displaystyle \angle A$ is parallel to the bisector of $\displaystyle \angle ABX.$
.Let $\displaystyle \theta \,=\,\angle ABX$

The slope of $\displaystyle BA$ is: .$\displaystyle m_1 \:=\:\tan\theta \:=\:\tfrac{3}{4} \quad\Rightarrow\quad \cos\theta \:=\:\tfrac{4}{5}$

The slope of $\displaystyle BD$ is: .$\displaystyle m_2 \:=\:\tan\tfrac{\theta}{2} \;=\;\sqrt{\frac{1-\cos\theta}{1 +\cos\theta}} \;=\;\sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}} \;=\;\sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}}$ .$\displaystyle = \;\sqrt{\frac{1}{9}} \;=\;\frac{1}{3}$

You have the slope of the angle bisector, $\displaystyle \tfrac{1}{3}$
. . and a point on the angle bisector $\displaystyle (4,3)$

You can now write its equation, right?