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Math Help - Bisector of an acute angle (Help)

  1. #1
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    Exclamation Bisector of an acute angle (Help)

    Find the equation of the line that bisects the acute angles determined by the given lines.

    1)  x-6y-12=0 and  4x-6y-9=0

    Attempt:

    d1=\frac{x-6y-12}{-\sqrt{37}} ,  \frac{4x-6y-9}{-2\sqrt{13}}
    So d1+d2 =0
    d1=\frac{x-6y-12}{-\sqrt{37}} +  \frac{4x-6y-9}{-2\sqrt{13}}


    =  \frac{(-4\sqrt{37} -2\sqrt{13} )x +(6\sqrt{37} +12\sqrt{13} )y+(-9\sqrt{37} -24\sqrt{13} )}{2\sqrt{481}}

    How to make the equation free of square roots?
    And is my solution correct ( if not please follow the procedure that i have done which is d1+d2=0 )
    Thank you.
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  2. #2
    Super Member malaygoel's Avatar
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    <br />
d1=\frac{x-6y-12}{-\sqrt{37}}<br />
, <br />
\frac{4x-6y-9}{-2\sqrt{13}}<br />
    why you have found distances from the origin?
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  3. #3
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    Quote Originally Posted by malaygoel View Post
    why you have found distances from the origin?
    Oh sorry it should be x prime and y prime.
    Please answer my question?
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by mj.alawami View Post
    Find the equation of the line that bisects the acute angles determined by the given lines.

    1)  x-6y-12=0 and  4x-6y-9=0

    Attempt:

    d1=\frac{x-6y-12}{-\sqrt{37}} ,  \frac{4x-6y-9}{-2\sqrt{13}}
    It is
    d1=\frac{|x-6y-12|}{\sqrt{37}} ,  \frac{|4x-6y-9|}{2\sqrt{13}}

    So d1+d2 =0
    so there are two equtions:
    d1+d2=0
    d1=d2
    d1=\frac{x-6y-12}{-\sqrt{37}} +  \frac{4x-6y-9}{-2\sqrt{13}}
    0=\frac{x-6y-12}{\sqrt{37}} +  \frac{4x-6y-9}{2\sqrt{13}}
    \frac{x-6y-12}{\sqrt{37}} =  \frac{4x-6y-9}{2\sqrt{13}}


    I have not checked calculations...will do in a short time.
    Last edited by malaygoel; June 26th 2009 at 11:11 PM.
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  5. #5
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    You have an error in your second equation it should be
    d1=-d2
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