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Math Help - Bisector

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    Bisector

    Find the equation of the bisector of the acute angle formed by the lines 3x-y-5=0 and 3x+4y-12=0.

    please help me and show the solution.. thanks!
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    Quote Originally Posted by lance View Post
    Find the equation of the bisector of the acute angle formed by the lines 3x-y-5=0 and 3x+4y-12=0.

    please help me and show the solution.. thanks!
    Write the equations of the lines in y = mx + c form...

    y = 3x - 5 and y = -\frac{3}{4}x + 3.

    Note that the gradient of any line m = \tan{\theta}, where \theta is the angle made with the positive x axis.

    So in the first case:

    \tan{\theta_1} = 3 and so \theta_1 = \arctan{3}.

    In the second case:

    \tan{\theta_2} = -\frac{3}{4}, so \theta_2 = \arctan{\left(-\frac{3}{4}\right)}.

    The angle between them is

    \theta = \frac{\theta_1 + \theta_2}{2}.

    Therefore the gradient of the line is m = \tan{\theta}.


    Now, work out the point of intersection of the two lines, and you can substitute this point into y = mx + c to find c.
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