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Equations of a triangle

  1. Apr 12, 2014 #1
    Hi,
    I am working on some matlab for my engineering classes and I need help with understanding a given diagram.

    In the attached diagram, I need help understanding how we determined the equations

    For example, I would like to know the equations for a triangle with the peak at 2/3L

    I know the equations would be from

    0<=x<=2/3L
    2/3L<=x<=L

    but I need help creating the equations.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Apr 12, 2014 #2
    If you have two points on a straight line at (0,0) and (L/2,h) and you are asked to determine the equation of that line, do you know how to determine it, say, in slope-intercept form?

    Same thing for the other line through the points (L/2,h) and (L,0).

    Chet
     
  4. Apr 12, 2014 #3
    Okay that makes sense, but when I solve like that I end up with a negative in the second equation and still not sure where the x comes in
    m1=h/L/2 or 2h/L
    m2=-h/L/2 or -2h/L

    I apply that to my 2/3 triangle

    m1=3h/2L
    m2=-3h/L
    both my m2's are negative but the given equation has the original m2 as positive.

    btw thanks for the tip so far

    EDIT:

    actually give me a minute before you answer, i think I need to plug it into y=mx+b
     
    Last edited: Apr 12, 2014
  5. Apr 12, 2014 #4
    So my actual problem has the triangle peak at 3/4L and the peak is represented by h


    please verify my equations
    slope 1 = 4h/3L
    slope 2 = -4h/L

    equation 1 = 4hx/3L
    equation 2 = -4hx/L + 12h or (4h(3L-x))/L

    Please verify I did it correctly

    Thank you
     
  6. Apr 12, 2014 #5
    You can verify it yourself. In eqn 2, if you substitute x = L, do you get y = 0?

    Chet
     
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