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Math Help - Finding a point position on a shape after the shape was rotated

  1. #1
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    Finding a point position on a shape after the shape was rotated

    Hi. First I'd like to let you know that I'm probably in the wrong section, but this is because I really, honestly, don't know where to put this question. If I knew what the problem is about, I'd have looked up on the Internet before asking.

    So, I've made a picture explaining my problem : http://i.imgur.com/EkQZq.png .

    If anyone can explain me how to solve this kind of problem and tell me what domain it is related to, it would be greatly appreciated.

    Thanks already.
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  2. #2
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    Re: Finding a point position on a shape after the shape was rotated

    This is a standard problem if you allow some changes.
    First lets say the center is (0,0); that is your x=0~\&~y=0.
    Then if the circle most have radius of say five.
    The ‘red’ point (0-4,0-4)=(-4,-4) has well-known coordinates:
    \left( {4\sqrt 2 \cos \left( {\frac{{5\pi }}{4} + t} \right),4\sqrt 2 \sin \left( {\frac{{5\pi }}{4} + t} \right)} \right),\,t = 0

    If you let t=\frac{-\pi}{2} then that rotates the point (-4,-4) to the point (-4,4).

    Does that help?
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  3. #3
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    Re: Finding a point position on a shape after the shape was rotated

    Hello, Minikloon!

    Another solution . . .


    Say I have a circle with center C(x_o,y_o).
    A point P is at (x_o\!-\!4,y_o\!-\!4).
    Rotate the circle \theta radians counterclockwise.
    Determine the new position of point P.

    Code:
                  * * *
              *           *
            *               *
           *                 *
                 
          *      4  C         *
          *   + - - o         *
          *   :   *           *
             4: *
           *  o              *
            * P             *
              *           *
                  * * *

    Point P lies on the circle with center C(x_o,y_o) and radius CP = 4\sqrt{2}.

    The coordinates of P are: . \begin{Bmatrix}x &=& x_o + 4\sqrt{2}\cos\left(\theta + \frac{5\pi}{4}\right) \\ \\[-4mm] y &=& y_o + 4\sqrt{2}\sin\left(\theta + \frac{5\pi}{4}\right) \end{Bmatrix}

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  4. #4
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    Re: Finding a point position on a shape after the shape was rotated

    Quote Originally Posted by Soroban View Post
    Hello, Minikloon!
    Another solution . . .
    Another solution???
    Come on, that is identical to the solution that I posted. Is it not?
    Please correct me if I am wrong. How is it another solution?
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  5. #5
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    Re: Finding a point position on a shape after the shape was rotated

    Hello, Plato!

    Don't get your shorts in a knot . . . identical?

    You made some changes . . . placing the center at the origin.
    . . I didn't ... I solved it using the original information.

    I believe that even the touchy moderators will agree
    . . that I made a significant contribution.

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