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Math Help - Bisector of an acute angle Q2 (Help)

  1. #1
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    Exclamation Bisector of an acute angle Q2 (Help)

    [FONT="Arial"]Find the equation of the bisector of the acute angle whose equation is
     4x-5y-10 and  x-4y+6

    Attempt

    L1=  \frac{4x'-5y'-10}{-\sqrt{41}}

    L2=  \frac{x'-4y'+6}{-\sqrt{17}}

    Since, the bisect line is in the middle of the lines then we use the formula \boxed{ -L2=L1 }

     -[-\frac{x-4y+6}{\sqrt{17}}]=-[\frac{4x-5y-10}{\sqrt{41}}]



    =  \boxed{(\sqrt{41}-4\sqrt{17})x+(-4\sqrt{41}-5\sqrt{17})y+(6\sqrt{41}-10\sqrt{17})=0}

    Is my answer correct?
    Thank you.
    Last edited by mj.alawami; June 27th 2009 at 12:46 AM.
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    [FONT="Arial"]Find the equation of the bisector of the acute angle whose equation is
     4x-5y-10 \bold{\color{red}= 0} and  x-4y+6\bold{\color{red}= 0} <<<<<< I assume that you mean these equations

    Attempt

    L1=  \frac{4x'-5y'-10}{-\sqrt{41}}

    L2=  \frac{x'-4y'+6}{-\sqrt{17}}

    Since, the bisect line is in the middle of the lines then we use the formula \boxed{ -L2=L1 }

     -[-\frac{x-4y+6}{\sqrt{17}}]=-[\frac{4x-5y-10}{\sqrt{41}}]



    =  \boxed{(\sqrt{41}-4\sqrt{17})x+(-4\sqrt{41}-5\sqrt{17})y+(6\sqrt{41}-10\sqrt{17})=0}

    Is my answer correct?
    Thank you.
    I've drawn the three lines and found out that your bisector doesn't pass through the point of intersection.

    I've got as the bisector the line with the following equation:

    \left(\dfrac{4 \cdot \sqrt{41}}{41} + \dfrac{\sqrt{17}}{17} \right)x  - \left(\dfrac{5 \cdot \sqrt{41}}{41} + \dfrac{4 \cdot \sqrt{17}}{17}\right)y + \left(\dfrac{6\cdot \sqrt{17}}{17} - \dfrac{10 \cdot \sqrt{41}}{41} \right) = 0

    I've attached the graph of my solution.
    Attached Thumbnails Attached Thumbnails Bisector of an acute angle Q2 (Help)-winkhalbneu2ger.png  
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  3. #3
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    Quote Originally Posted by earboth View Post
    I've drawn the three lines and found out that your bisector doesn't pass through the point of intersection.

    I've got as the bisector the line with the following equation:

    \left(\dfrac{4 \cdot \sqrt{41}}{41} + \dfrac{\sqrt{17}}{17} \right)x  - \left(\dfrac{5 \cdot \sqrt{41}}{41} + \dfrac{4 \cdot \sqrt{17}}{17}\right)y + \left(\dfrac{6\cdot \sqrt{17}}{17} - \dfrac{10 \cdot \sqrt{41}}{41} \right) = 0

    I've attached the graph of my solution.

    So is my answer wrong?
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  4. #4
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    Quote Originally Posted by mj.alawami View Post
    [FONT="Arial"]...

     -[-\frac{x-4y+6}{\sqrt{17}}]=-[\frac{4x-5y-10}{\sqrt{41}}]



    ...
    I take this equation:

    <br />
-\left[-\frac{x-4y+6}{\sqrt{17}}\right]=-\left[\frac{4x-5y-10}{\sqrt{41}}\right]<br />

    <br />
\left[-\frac{x-4y+6}{\sqrt{17}}\right]=\left[\frac{4x-5y-10}{\sqrt{41}}\right]<br />

    <br />
-(x-4y+6) \cdot \sqrt{41}=(4x-5y-10) \cdot \sqrt{17}<br />

    <br />
(-x+4y-6) \cdot \sqrt{41}=(4x-5y-10) \cdot \sqrt{17}<br />

    <br />
-x\cdot \sqrt{41}+4\cdot \sqrt{41}y-6\cdot \sqrt{41}=4\cdot \sqrt{17}<br />
x-5\cdot \sqrt{17}<br />
y-10\cdot \sqrt{17}<br /> <br />

    <br />
0=(4\cdot \sqrt{17} + \sqrt{41})x-(5\cdot \sqrt{17}+4\cdot \sqrt{41})y+(6\cdot \sqrt{41}-10\cdot \sqrt{17})<br /> <br />

    By comparison you'll see that you have made a tiny sign mistake in the first bracket.
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  5. #5
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    Hello, mj.alawami!

    Your game plan is correct,
    but you have some incorrect minus-signs.


    Find the equation of the bisector of the acute angle whose equation is:
    . .  4x-5y-10 \:=\:0 and  x-4y+6 \:=\:0


    Attempt

    L_1 \:=\:\frac{4x-5y-10}{{\color{red}-}\sqrt{41}}\qquad L_2\:=\:\frac{x'-4y'+6}{{\color{red}-}\sqrt{17}} . ??
    There is no minus-sign in the denominator.

    Since, the bisect line is in the middle of the lines,
    then we use the formula:  {\color{red}-}L2\,=\,L1 . ??
    The distances are of opposite signs?

     -\left[-\frac{x-4y+6}{\sqrt{17}}\right]\;=\;-\left[\frac{4x-5y-10}{\sqrt{41}}\right]

    =  \boxed{(\sqrt{41}-4\sqrt{17})x+(-4\sqrt{41}-5\sqrt{17})y+(6\sqrt{41}-10\sqrt{17})=0}

    Is my answer correct? .
    . . . Sorry, no

    But you were close . . .

    Try it again without those extra minus-signs.

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  6. #6
    Senior Member pankaj's Avatar
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    Here is a general method which I have been using.
    Let the two lines be
    a_{1}x+b_{1}y+c_{1}=0
    a_{2}x+b_{2}y+c_{2}=0

    Remember to write the two equations in such a way that c_{1}>0 and c_{2}>0.

    The eqution of the bisectors are

     <br />
\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}}  =\pm<br />
\frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}<br />

    If a_{1}a_{2}+b_{1}b_{2}>0,then use of + sign in above equation will give OBTUSE angle bisector

    i.e.
     <br />
\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}}  =+<br />
\frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}<br />

    is eqution of OBTUSE angle bisector and

     <br />
\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}}  =-<br />
\frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}<br />
    gives equation of ACUTE angle bisector
    .................................................. .................................................. ..
    If a_{1}a_{2}+b_{1}b_{2}<0,then use of - sign in above equation will give OBTUSE angle bisector

    i.e.
     <br />
\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}}  =-<br />
\frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}<br />

    is eqution of OBTUSE angle bisector and

     <br />
\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}}  =+<br />
\frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}<br />
    is equation of ACUTE angle bisector
    .................................................. .............................................

    General rule
    (i)  <br />
\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^2+b_{1}^2}}  =\pm<br />
\frac{a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^2+b_{2}^2}}<br />
    where, c_{1}>0,c_{2}>0

    (ii)Find sign of a_{1}a_{2}+b_{1}b_{2}.

    (iii)The use of this sign always gives the OBTUSE angle bisector and the use of other sign gives ACUTE angle bisector
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