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Math Help - Analytical Geometry

  1. #1
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    Analytical Geometry

    Determine the equation of the straight bisector of the acute angle that the straight y = 2x form with the axis x.
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  2. #2
    Member billa's Avatar
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    I have never done a problem like this but I will give it a try.

    The angle between the x-axis and y=2x is
    atan(opp/adj)=atan(rise/run)=atan(m)=atan(2)=1.1071 radians
    The bisector of an angle make half this angle with the x-axis, so .55355 radians
    tan(.55355) = opp/adj = rise/run = m2 = .6180
    So I think the formula for a bisector is y=.6180*x or y=(tan(atan(angle)/2))*x
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  3. #3
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    Hello, Apprentice123!

    We can solve this without a calculator
    . . which is what they probably expected.



    Determine the equation of the bisector of the acute angle
    formed by the straight line y \:=\: 2x and the x-axis.
    Code:
            |      A
            |     /          B
            |    /         *
            |   /       *
            |  / θ  *
            | /   *
            |/ *  θ
          O * - - - - - - - X
            |

    Line OA has the equation: . y = 2x
    . . Let \theta \,=\,\angle AOX.

    Its slope is: . m_1 \:=\:2\:=\:\tan \theta

    We want: . m_2 \:=\:\tan\left(\tfrac{\theta}{2}\right)


    Since \tan\theta \:=\:\frac{2}{1} \:=\:\frac{opp}{adj}, \theta is in a right triangle with: opp = x,\;adj = 1
    And Pythagorus tell us that: . hyp = \sqrt{1^2+2^2} \:=\:\sqrt{5}
    . . Hence: . \cos\theta \:=\:\frac{adj}{hyp}\:=\:\frac{1}{\sqrt{5}}


    We have: . \tan\tfrac{\theta}{2} \;=\;\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}  {2}} \;=\;\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \:= . \sqrt{\frac{1-\frac{1}{\sqrt{5}}}{1 + \frac{1}{\sqrt{5}}}} \:=\:\sqrt{\frac{\sqrt{5}-1}{\sqrt{5}+1}}

    Rationalize: . m_2 \;=\;\sqrt{\frac{\sqrt{5}-1}{\sqrt{5}+1}\cdot\frac{\sqrt{5}-1}{\sqrt{5}-1}}\:=\:\sqrt{\frac{(\sqrt{5}-1)^2}{4}} \:=\:\frac{\sqrt{5}-1}{2}


    Therefore, the equation of the bisector is: . y \;=\;\left(\frac{\sqrt{5}-1}{2}\right)x

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Apprentice123!

    We can solve this without a calculator
    . . which is what they probably expected.



    Code:
            |      A
            |     /          B
            |    /         *
            |   /       *
            |  / θ  *
            | /   *
            |/ *  θ
          O * - - - - - - - X
            |

    Line OA has the equation: . y = 2x
    . . Let \theta \,=\,\angle AOX.

    Its slope is: . m_1 \:=\:2\:=\:\tan \theta

    We want: . m_2 \:=\:\tan\left(\tfrac{\theta}{2}\right)


    Since \tan\theta \:=\:\frac{2}{1} \:=\:\frac{opp}{adj}, \theta is in a right triangle with: opp = x,\;adj = 1
    And Pythagorus tell us that: . hyp = \sqrt{1^2+2^2} \:=\:\sqrt{5}
    . . Hence: . \cos\theta \:=\:\frac{adj}{hyp}\:=\:\frac{1}{\sqrt{5}}


    We have: . \tan\tfrac{\theta}{2} \;=\;\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}  {2}} \;=\;\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \:= . \sqrt{\frac{1-\frac{1}{\sqrt{5}}}{1 + \frac{1}{\sqrt{5}}}} \:=\:\sqrt{\frac{\sqrt{5}-1}{\sqrt{5}+1}}

    Rationalize: . m_2 \;=\;\sqrt{\frac{\sqrt{5}-1}{\sqrt{5}+1}\cdot\frac{\sqrt{5}-1}{\sqrt{5}-1}}\:=\:\sqrt{\frac{(\sqrt{5}-1)^2}{4}} \:=\:\frac{\sqrt{5}-1}{2}


    Therefore, the equation of the bisector is: . y \;=\;\left(\frac{\sqrt{5}-1}{2}\right)x

    ok. Thank you
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