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
A+r(B-A)=C+s(D-C), or
Ax+r(Bx-Ax)=Cx+s(Dx-Cx)
Ay+r(By-Ay)=Cy+s(Dy-Cy) for some r,s in [0,1]
Solving the above for r and s yields
(Ay-Cy)(Dx-Cx)-(Ax-Cx)(Dy-Cy)
r = ----------------------------- (eqn 1)
(Bx-Ax)(Dy-Cy)-(By-Ay)(Dx-Cx)
(Ay-Cy)(Bx-Ax)-(Ax-Cx)(By-Ay)
s = ----------------------------- (eqn 2)
(Bx-Ax)(Dy-Cy)-(By-Ay)(Dx-Cx)
Let P be the position vector of the intersection point, then
P=A+r(B-A) or
Px=Ax+r(Bx-Ax)
Py=Ay+r(By-Ay)
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.