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Foot of perpendicular

  1. Sep 25, 2012 #1
    I apologise in advance for an extremely trivial question, i am returning to brush up on some basic maths and cant seem to arrive at the answer

    Find the coordinates of the foot of the perpendicular from the point A with coordinates (-5,5) to the line 10x+2y-3= 0

    1.) gradient of original line -10/2, gradient of intercepting line 1/5
    2.) equation of intercepting line y-5 = 1/5(x+1) *5
    3.)5y-25 = x+1 re-arranging gives 5y-x = 26
    4.) solve simultaneously -10x-50y=-260 ( *-10 )
    10x+2y=3

    -48y=-257

    y=5.354
    Substituting y into either equation
    x= -0.7708

    From the program i know these answers arnt correct

    Any pointers would be appreciated,

    Thanks
     
  2. jcsd
  3. Sep 25, 2012 #2

    Mark44

    Staff: Mentor

    Mistake above. The known point on the intersecting line is (-5, 5), so the equation of this line would be y - 5 = (1/5)(x - (-5))

    Why do you have "*5" on the right side of your equation?
     
  4. Sep 25, 2012 #3
    Hi,

    I had the *5 to remove the fraction, dont know how i ended up with the 1 on the right hand side. That sorted it. Thanks for spotting that, such a silly mistake. Wouldnt think id been staring at it for half an hour. Probably suggests i need some sleep :bugeye:
     
  5. Sep 25, 2012 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Check the red part.

    ehild
     
  6. Sep 25, 2012 #5

    Mark44

    Staff: Mentor

    If you multiply one side by 5, you need to also multiply the other side by the same number. Maybe you did this, but it doesn't show in your work.

    It's best to start a new line when you do this kind of operation.
     
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