# Line Segment Intersection

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incidenta5    122
I've been scratching my head over this problem for 2 days now.. so I decided to post. Given the line segments (-4,3) to (2,-3) and (-2,-4) to (6,-1) do they intersect and where? (done in parametric form). I have the following:
x(a) = -4 + 6a
y(a) = 3 - 6a

x1(a) = -2 + 8a
y1(a) = -4 + 3a

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x(a) = x1(a)
-2a = 2
a = -1

y(a) = y1(a)
7 = 9a
a = 7/9

My question is this: If I get a negative alpha for x(a) = x1(a) then don't these line segments not intersect? I graphed it and it looks like they do intersect.... Any thoughts?

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eriatarka    122
Your problem is that you use the same parameter for both line segments; they will typically have different parameters at their intersection point.

Use x(a), y(a) for the first segment and x1(b), y1(b) for the second one and repeat your calculations.

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apatriarca    2365
Your equations are wrong. The segments in parametric form are:

P(t) = (-4, 3) + t*(6, 6)
Q(s) = (-2,-4) + s*(8, 3)

You want to find the values of t and s where P(t) = Q(s) so you have to solve the following system of equations in t and s:

-4 + 6t = -2 + 8s
3 - 6t = -4 + 3s

6t - 8s = 2
6t + 3s = 7

Solve this system to find s and t. If these values are in [0,1] than the segments intersect. To find the point of intersection you can than substitute the value of s or t in the formula of the correct segment.

P.S. The solution should be t = 31/33, s = 5/11 and the two segments intersect.

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incidenta5    122
Thanks for pointing that out.