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Math Help - Help with equation of the bisector of an angle in a triangle

  1. #1
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    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.
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  2. #2
    MHF Contributor red_dog's Avatar
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    The equation of the side AB: 3x-4y=0

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

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

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

    We get two equations: 2x+4y-15=0 and 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.
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  3. #3
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    Here is a hint.
    If \overrightarrow a \;\& \,\overrightarrow b are vectors then the bisector of the angle between them is \left\| {\overrightarrow b } \right\|\overrightarrow a  + \left\|{\overrightarrow a } \right\|\overrightarrow b .
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  4. #4
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    Hello, ninni!

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


    Find the equation of the bisector of angle A of \Delta ABC.
    The coordinates of the vertices are: . 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 BD = bisector of \angle ABX.

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

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

    The slope of BD is: . 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}}} . = \;\sqrt{\frac{1}{9}} \;=\;\frac{1}{3}


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

    You can now write its equation, right?

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