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Math Help - Analytic geometry help?

  1. #1
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    Exclamation Analytic geometry help?

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

     2x+3y-5=0 and  x+6y-8=0

    Attempt:
    I know that i have to get the distance of d1 and d2 and then apply the equation
    d1+d2=0

    But when i try doing it i get very complex root forms and i guess it is wrong so how to do it ? :'(
    Last edited by mj.alawami; June 26th 2009 at 11:50 AM.
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Find the equation of that line that bisects the acute angles determined by the given lines.

     2x+3y-5=0 and x+6y-8=0

    Attempt:
    I know that i have to get the distance of d1 and d2 and then apply the equation
    d1+d2=0

    But when i try doing it i get very complex root forms and i guess it is wrong so how to do it ? :'(
    1. Calculate the intersection point. I've got P\left(\frac23\ ,\ \frac{11}9\right)

    2. Now transform the given equations such that the direction vector (or in your case the normal vector to the direction) is a unit vector:

    2x+3y-5=0~\implies~\frac2{\sqrt{13}}x+\frac3{\sqrt{13}}y-\frac5{\sqrt{13}}=0

    x+6y-8=0~\implies~\frac1{\sqrt{37}}x+\frac6{\sqrt{37}}y-\frac8{\sqrt{37}}=0

    3. The angle bisector passes through P and have the direction:
    <br />
\left(\frac2{\sqrt{13}} + \frac1{\sqrt{37}}  \right)x + \left(\frac3{\sqrt{13}} + \frac6{\sqrt{37}}  \right)y - C =0

    Plug in the coordinates of C to complete the equation of the line.
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  3. #3
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    Quote Originally Posted by earboth View Post
    1. Calculate the intersection point. I've got P\left(\frac23\ ,\ \frac{11}9\right)

    2. Now transform the given equations such that the direction vector (or in your case the normal vector to the direction) is a unit vector:

    2x+3y-5=0~\implies~\frac2{\sqrt{13}}x+\frac3{\sqrt{13}}y-\frac5{\sqrt{13}}=0

    x+6y-8=0~\implies~\frac1{\sqrt{37}}x+\frac6{\sqrt{37}}y-\frac8{\sqrt{37}}=0

    3. The angle bisector passes through P and have the direction:
    <br />
\left(\frac2{\sqrt{13}} + \frac1{\sqrt{37}}  \right)x + \left(\frac3{\sqrt{13}} + \frac6{\sqrt{37}}  \right)y - C =0

    Plug in the coordinates of C to complete the equation of the line.

    I am getting different answers :

     \frac{13+2\sqrt{481}}{13} x +\frac{78+2\sqrt{481}}{13}y-\frac{104-5\sqrt{481}}{13}=0
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  4. #4
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    Quote Originally Posted by mj.alawami View Post
    I am getting different answers :

     \frac{13+2\sqrt{481}}{13} x +\frac{78+2\sqrt{481}}{13}y-\frac{104-5\sqrt{481}}{13}=0
    I have attached a screenshot. Obviously we both have made a mistake somewhere.

    If (and when) I find my mistake I'll be back.
    Attached Thumbnails Attached Thumbnails Analytic geometry help?-winkhalb2geraden.png  
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  5. #5
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    Back again.

    I've attached my complete calculations. The graph shows the equations #1, #2 and #11
    Attached Thumbnails Attached Thumbnails Analytic geometry help?-ber_winkhalb2ger.png   Analytic geometry help?-graph_winkhalb2ger.png  
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  6. #6
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    Quote Originally Posted by earboth View Post
    Back again.

    I've attached my complete calculations. The graph shows the equations #1, #2 and #11

    I guess your answer is wrong because I am getting the answer as :

     \boxed {(-\sqrt{13}-2\sqrt{37})x+(-6\sqrt{13}-3\sqrt{37})y+(8\sqrt{15}+5\sqrt{37}}
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  7. #7
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    equation of angle bisector

    posted by mj.alawami and earboth

    This is an interesting problem which may still be looking for an answer

    I get the following

    y=-.3955x +1.486

    Method used employs trig to convert slopes to elevation angles.The angle for the bisector is the angle of lower line plus half the angle between the lines 21.58 degrees slope -.3955. line equation from point slope formula using the point (2/3,11/9)



    bj
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  8. #8
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    Quote Originally Posted by bjhopper View Post
    posted by mj.alawami and earboth

    This is an interesting problem which may still be looking for an answer ..... No

    I get the following

    y=-.3955x +1.486

    ...
    The result from my previous post was:

    \left(\dfrac{2\cdot \sqrt{13}}{13} + \dfrac{\sqrt{37}}{37}\right) \cdot x + \left(\dfrac{3 \cdot \sqrt{13}}{13} + \dfrac{6 \cdot \sqrt{37}}{37}\right) \cdot y -  \left(\dfrac{5\cdot \sqrt{13}}{13} + \dfrac{8 \cdot \sqrt{37}}{37}\right) = 0

    If you solve this equation for y and use approximate values you'll get:

    y =  - 0.3954474799\cdot x + 1.485853875
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