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Geometry help on rotations.

  1. Aug 7, 2014 #1
    1. The problem statement, all variables and given/known data

    1. A group of American physicists works on a project where planar lines are in the form
    X=t⋅P+s⋅Q
    where P , 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=t⋅P+s⋅Q in the following 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.

    2. Relevant equations
    I know for number 1, you basically just add the vector B.
    for 2 and 3 I do not know whether to use point slope form and just change the slope or if I need to change the coordinates to (-y,x) for 90 degree rotation and (-x,-y) for 180 degree rotation but those are for rotation about the origin and my problem does not state that. Does the slope for a 180 degree rotation go back to the same slope? I am really confused on which direction to take.

    3. The attempt at a solution
    attempt at part 1:


    2. Relevant equations



    3. The attempt at a solution
    Let a$, $b$, and $c$ be fixed reals satisfying $a^2+b^2\ne 0$.
    They need to know formulae for the images of the line $$a\cdot x+b\cdot y+c=0$$\\
    1. Under the translation by a vector $$B=[u,v]$$,\\
    First, solve for $y$ $\to$ $$y=\frac{-a}{b}\cdotx -\frac{c}{b}$$.\\
    So, we know that the line crosses the y-axis at $$-\frac{c}{b}$$ $\to$ $$(0, -\frac{c}{b}$$.\\
    The translation by a vector means that we add $B=[u,v]$,to yield\\
    $$(0, -\frac{c}{b}) + (u,v)=(u,v-\frac{c}{b})$$ $\to$ $$(u,\frac{bv-c}{b})$$\\
    Now we let $x_0=u, y_0=(\frac{bv-c}{b})$\\
    $$y-y_0=m(x-x_0) \to y-(\frac{bv-c}{b})=mx – mu$$\\
    $$-m\cotx + y = \frac{bv-c}{b}) –mu \to -m\cotx + y - \frac{bv-c}{b})+mu=0$$\\
     
  2. jcsd
  3. Aug 7, 2014 #2

    Simon Bridge

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    X=tP+sQ is actually a plane, hence the restriction that t+s=1 ... so you can rewrite:

    X=tP+(1-t)Q = Q+(P-Q)t ... which is a more familiar form of the line.

    Basically P and Q are both on the line, and P-Q is a vector pointing along the line.
    You know how to rotate a vector?

    I'd be thinking in terms of the vector Q-C ... may help to draw a sketch.
     
  4. Aug 8, 2014 #3

    Fredrik

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    You should use the preview feature to see what your post will look like before you submit it. You need to use ## instead of $. Using \\ to get a new line only works in certain environments, e.g. the align environment. It's \cdot x and \cot x, not \cdotx and \cotx.

    See the FAQ post on LaTeX for more information. https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
     
  5. Aug 8, 2014 #4

    Simon Bridge

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    Yeah - that would be:

    ------------------------------------------------------

    Let ##a##, ##b##, and ##c## be fixed reals satisfying ##a^2+b^2\neq 0##.
    They need to know formulae for the images of the line $$a\cdot x+b\cdot y+c=0$$
    1. Under the translation by a vector ##B=[u,v]##,
    First, solve for $$y \to y=\frac{-a}{b}\cdot x -\frac{c}{b}$$
    So, we know that the line crosses the y-axis at $$-\frac{c}{b} \to (0, -\frac{c}{b}$$.
    The translation by a vector means that we add ##B=[u,v]##, to yield
    $$(0, -\frac{c}{b}) + (u,v)=(u,v-\frac{c}{b})\to (u,\frac{bv-c}{b})$$
    Now we let ##x_0=u, y_0=(\frac{bv-c}{b})##
    $$y-y_0=m(x-x_0) \to y-(\frac{bv-c}{b})=mx – mu$$
    $$-m\cot x + y = \frac{bv-c}{b}) –mu \to -m\cot x + y - \frac{bv-c}{b})+mu=0$$
     
  6. Aug 9, 2014 #5
    I answered something similar you posted on the pre-calculus homework board.

    I'm confused by what you have posted here because it looks similar. The problem statement uses different variables. Yet your solution uses the variables from the other problem. I don't understand why you moved from the variables P, Q, s, and t to a, b, c, and y. Also, why has the condition changed from s + t = 1 to ##a^2 + b^2 \neq 0##
     
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