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Thread: Angular bisector

  1. #1
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    Angular bisector

    How to find eqn. of the bisector of angle between 2 lines which contains a given point.
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  2. #2
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    Re: Angular bisector

    Quote Originally Posted by mak29 View Post
    How to find eqn. of the bisector of angle between 2 lines which contains a given point.
    This question is far too vague to answer.
    There is no bisector unless the lines intersect. Do they intersect?
    Next if they intersect then there are two angles formed and so there are two bisectors.
    But each bisector is unique thus it makes no sense to say that a bisector contains a given point.

    So please repost a meaningful question.
    Thanks from topsquark
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  3. #3
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    Re: Angular bisector

    Let's say the lines are given by y= 0 and y= ax (We can always translate the coordinate system so that the point of intersection is (0, 0) and rotate the coordinate system so the x-axis lies along the first line). The line of bisection is of the form y= bx for some number b. Of course the "slope" of the line is the tangent of the angle so $\displaystyle b=tan(\theta/2)$ where $\displaystyle \theta= arctan(a)$.

    Now use a "half-angle formula": $\displaystyle tan(\theta/2)= \frac{tan(\theta)}{sec(\theta)+ 1}$. Here, of course, $\displaystyle tan(\theta)= a$ and $\displaystyle sec(\theta)= \sqrt{tan^2(\theta)+ 1}= \sqrt{a^2+ 1}$. So $\displaystyle b= \frac{a}{\sqrt{a^2+ 1}+ 1}$.
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