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.

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.

Source (copy-paste): http://www.gamers.org/dEngine/rsc/usenet/comp.graphics.algorithms.faq