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Math Help - Analytic Geometry Q11

  1. #1
    Member looi76's Avatar
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    Analytic Geometry Q11

    Question:
    Find the equation of the line that bisects the acute angles determined by the given lines: x-6y-12=0 and 4x-6y-9=0

    Attempt:

    L_1 \rightarrow x-6y-12=0


    d_1 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{x'-6y'-12}{\sqrt{1^2+6^2}} = \frac{x'-6y'-12}{+\sqrt{37}}



    L_2 \rightarrow 4x-6y-9=0


    d_2 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{4x'-6y'-9}{\sqrt{4^2+6^2}} = \frac{4x'-6y'-9}{-2\sqrt{13}}



    Since, the bisect line is in the middle of the lines then we use the formula \boxed{d_1+d_2=0}

    \left[\frac{x'-6y'-12}{+\sqrt{37}} + \frac{4x'-6y'-9}{-2\sqrt{13}} = 0\right]\sqrt{37}\times -2\sqrt{13}

    -2\sqrt{13}x' +12\sqrt{13}y' +24\sqrt{13} + 4\sqrt{37}x' - 6\sqrt{37}y' -9\sqrt{37} = 0

    Answer:
    \boxed{(-2\sqrt{13} + 4\sqrt{37})x' + (12\sqrt{13} - 6\sqrt{37})y' + (24\sqrt{13} - 9\sqrt{37}) = 0}

    Is my answer correct?
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  2. #2
    Super Member
    earboth's Avatar
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    Quote Originally Posted by looi76 View Post
    ...

    Is my answer correct?
    Yes.

    I've attached a graph of the three lines.
    Attached Thumbnails Attached Thumbnails Analytic Geometry Q11-winkhalb2.png  
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