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?

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.

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.

Thanks for pointing that out.