1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

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

    [​IMG]

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?