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Intersection of two lines

  1. Oct 14, 2004 #1
    I'm using the following equation to represent lines

    (x, y) = (start_x, start_y) + t((end_x, end_y) - (start_x, start_y))

    I'm trying to find the interesection point of two lines written in this form.

    I have been able to solve for t and plug it back into the equation, but i get two values of t when i solve for it. So that gives me 2 possible x values and 2 possible y values in the end. One x value is correct, and one y value is correct, but I get a wrong x value and a wrong y value.

    I'm wondering if theres a totally different way, or a way to get rid of the x and y value that I don't need.

  2. jcsd
  3. Oct 14, 2004 #2


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    Welcome to PF!
    You are going about this in a fundamentally wrong way.
    The parameter "t" need not have the same value at the point of intersection for each line!!
    What you need, is to solve a 2*2 system of two unknowns, the two parametrization variables:
    Let superscript "1" denote Line 1 expression, "2" Line 2 expression.
    Then we may write:
    [tex](x,y)=(x^{1}_{0},y^{1}_{0})+t^{1}(x_{1}^{1}-x_{0}^{1}, y_{1}^{1}-y_{0}^{1})[/tex]
    And line 2:
    [tex](x,y)=(x^{2}_{0},y^{2}_{0})+t^{2}(x_{1}^{2}-x_{0}^{2}, y_{1}^{2}-y_{0}^{2})[/tex]
    Set these expressions equal to each other, and solve for [tex]t^{1},t^{2}[/tex]

    You may also eliminate the parametrization variable, and solve the following system
    for x and y:
  4. Oct 14, 2004 #3
    Ok, thanks.

    What I ended up doing was turning the (x, y) = (a, b) + t(c - a, d - b) into y = mx + b. Then I solved for x and y.

    The reason I wanted t was to know whether the intersection was actually between the two points (0 <= t <= 1). But I can just plug the point I found back into the first equation and solve for t.

    x = a + t(c - a)
    t = (x - a) / (c - a)
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