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Find the formula for X=tP+sQ under a translation and a rotation

  1. Aug 7, 2014 #1
    1. The problem statement, all variables and given/known data
    A group of American physicist works on a project where planar lines are in the form X=tP+sQ, where P and Q are two fixed different points and s, t are varying reals satisfying s+t=1. They need to know formulae for the images of the line X=tP+sQ in the following three cases:
    1. Under the translation by a vector B
    2. Under rotation about a point C by 180 degrees
    3. Under rotation about a point C by 90 degrees
    Please provide those formulae and a justification for them.


    2. Relevant equations
    X=tP+sQ
    s+t=1


    3. The attempt at a solution
    A translation by a vector, B will preserve length and slope, so the new formulae is X=tP+sQ+B.

    A rotation will preserve length, but not necessarily slope. I know that with points, a 90 degree rotation will give (x,y) -> (y,-x) and that a rotation of 180 degrees will give (x,y) -> (-x,y).

    I'm not sure how to use this information to get to a clear answer, with a formula and justification for it.
     
  2. jcsd
  3. Aug 7, 2014 #2

    Ray Vickson

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    There are standard formulas for the transition from ##(x,y)## to ##(x',y')## under a rotation through angle ##\theta##. Just use them on each of the points P and Q.
     
  4. Aug 7, 2014 #3
    Where would I find these formulas? I have tried google and come up with nothing...?
     
  5. Aug 8, 2014 #4

    verty

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    For a rotation at C, translate C to the origin, rotate, translate the origin to C.
     
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