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Need help with a coordinate equation

  1. Jan 6, 2009 #1
    See the attached image first. Given that I know points P1 and P2, what would the equation be to get point P3, given that the angle is 90 degrees and the distance from P2 to P3 is 1/5 the distance from P1 to P2?

    Thanks in advance!!!

    Attached Files:

  2. jcsd
  3. Jan 6, 2009 #2


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    Welcome to PF!

    Hi chuyler1! Welcome to PF! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
  4. Jan 6, 2009 #3
    I haven't a clue where to begin. It's been 8 years since the last time I had to think about calculus. I found some equations on how to rotate a point by 90 degrees but the only examples are to rotate around (0,0).

    This calculation has to be performed hundreds of times in a short amount of time so if the equation is too complex i may just fudge it by increasing point P2's y by L/5. It simplifies the problem but doesn't provide desirable results in all situations.
  5. Jan 6, 2009 #4


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    Could you use the fact that the slopes of perpendicular lines are negative reciprocals of each other? I.e.:

    m1 = -1/m2
  6. Jan 7, 2009 #5
    What is m1 and m2 in your equation?
  7. Jan 7, 2009 #6


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    They are the slopes of the two lines you drew.

    m1 = slope of line P1-P2
    m2 = slope of line P2-P3
  8. Jan 7, 2009 #7


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    No calculus is needed here... just vector algebra.
    (Will P1-P2 ever be vertical?)
    Are you willing to use trig-functions?
  9. Jan 7, 2009 #8
    Yes, P1 and P2 are arbitrary, they could be any angle or even reversed. I have trig functions available to me so a solution involving them isn't a problem.
  10. Jan 7, 2009 #9


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    I've had some more time to think about this. We won't even need trig!


    In the figure, note that we form two similar right triangles, with a scale factor of 1:5.

    To get the x-coordinate of P3, subtract (y2-y1)/5 from x2.
    To get the y-coordinate of P3, add (x2-x1)/5 to y2.
  11. Jan 7, 2009 #10
    Awesome! Works perfectly!
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