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Thread: Analytical Geometry

  1. #1
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    Analytical Geometry

    Determine the equation of the straight bisector of the acute angle that the straight $\displaystyle y = 2x$ form with the axis $\displaystyle x$.
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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 $\displaystyle y \:=\: 2x$ and the $\displaystyle x$-axis.
    Code:
            |      A
            |     /          B
            |    /         *
            |   /       *
            |  / θ  *
            | /   *
            |/ *  θ
          O * - - - - - - - X
            |

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

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

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


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


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

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


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

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

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

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


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


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

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


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

    ok. Thank you
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