This problem can be extremely easy or extremely difficult; it depends on your application. If all you want is the intersection point, the following should work:
Let A,B,C,D be 2-space position vectors. Then the directed line segments AB & CD are given by:
AB=A+r(B-A), r in [0,1]
CD=C+s(D-C), s in [0,1]
If AB & CD intersect, then
Ay+r(By-Ay)=Cy+s(Dy-Cy) for some r,s in [0,1]
Solving the above for r and s yields
r = ----------------------------- (eqn 1)
s = ----------------------------- (eqn 2)
Let P be the position vector of the intersection point, then
By examining the values of r & s, you can also determine some other limiting conditions:
If 0<=r<=1 & 0<=s<=1, intersection exists
r<0 or r>1 or s<0 or s>1 line segments do not intersect
If the denominator in eqn 1 is zero, AB & CD are parallel
If the numerator in eqn 1 is also zero, AB & CD are collinear.
- If they are collinear, then the segments may be projected to the x- or y-axis, and overlap of the projected intervals checked.
- If the intersection point of the 2 lines are needed (lines in this context mean infinite lines) regardless whether the two line segments intersect, then
- If r>1, P is located on extension of AB
- If r<0, P is located on extension of BA
- If s>1, P is located on extension of CD
- If s<0, P is located on extension of DC
Also note that the denominators of eqn 1 & 2 are identical.
Source (copy-paste): http://www.gamers.org/dEngine/rsc/usenet/comp.graphics.algorithms.faq