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Math Help - Transformation geometry

  1. #1
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    Transformation geometry

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  2. #2
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    Hello walreinlim88
    Quote Originally Posted by walreinlim88 View Post
    I think there is a mistake in the question. I believe the transformation matrix should be:
    \begin{pmatrix}\dfrac{\sqrt5}{3}&\dfrac23\\ \dfrac23&-\dfrac{\sqrt5}{3}\end{pmatrix}
    which represents a reflection in the line through the origin that makes an anticlockwise angle \tfrac12\theta with the x-axis, where \cos\theta = \frac{\sqrt5}{3}.

    See the diagram attached, where PQ is the mirror-line. Note that A(1,0) \to A'(\sqrt5/3,2/3) and B(0,1) \to B'(2/3,-\sqrt5/3).


    It's fairly straighforward to show that if \cos\theta = \frac{\sqrt5}{3}, then \tan\tfrac12\theta = \frac{3-\sqrt5}{2}, and hence the equation of the mirror (invariant) line is
    y=\frac{3-\sqrt5}{2}x
    (Use \cos^2\tfrac12\theta = \tfrac12(1+\cos\theta) and \tan\tfrac12\theta  = \sqrt{\sec^2\tfrac12\theta -1}.)

    Can you continue from here?


    Grandad
    Attached Thumbnails Attached Thumbnails Transformation geometry-untitled.jpg  
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  3. #3
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    how about 2nd part? i dont know how to find the point of T.
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  4. #4
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    Hello walreinlim88
    Quote Originally Posted by walreinlim88 View Post
    how about 2nd part? i dont know how to find the point of T.
    Find the image of one of the points in the mirror-line; join its reflection to the other point. T is the point where this line intersects the mirror-line.

    Grandad
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