How do I determine the intersection point of two lines in GDI+?

Question

I\'m using .NET to make an application with a drawing surface, similar to Visio. The UI connects two objects on the screen with Graphics.DrawLine. This simple implementation wor

Answer 1

The representation of lines by y = mx + c is problematic for computer graphics, because vertical lines require m to be infinite.

Furthermore, lines in computer graphics have a start and end point, unlike mathematical lines which are infinite in extent. One is usually only interested in a crossing of lines if the crossing point lies on both the line segments in question.

If you have two line segments, one from vectors x1 to x1+v1, and one from vectors x2 to x2+v2, then define:

a = (v2.v2 v1.(x2-x1) - v1.v2 v2.(x2-x1)) / ((v1.v1)(v2.v2) - (v1.v2)^2)
b = (v1.v2 v1.(x2-x1) - v1.v1 v2.(x2-x1)) / ((v1.v1)(v2.v2) - (v1.v2)^2)

where for the vectors p=(px,py), q=(qx,qy), p.q is the dot product (px * qx + py * qy). First check if (v1.v1)(v2.v2) = (v1.v2)^2 - if so, the lines are parallel and do not cross.

If they are not parallel, then if 0<=a<=1 and 0<=b<=1, the intersection point lies on both of the line segments, and is given by the point

x1 + a * v1

Edit The derivation of the equations for a and b is as follows. The intersection point satisfies the vector equation

x1 + a*v1 = x2 + b*v2

By taking the dot product of this equation with v1, and with v2, we get two equations:

v1.v1*a - v2.v1*b = v1.(x2-x1)
v1.v2*a - v2.v2*b = v2.(x2-x1)

which form two linear equations for a and b. Solving this system (by multiplying the first equation by v2.v2 and the second by v1.v1 and subtracting, or otherwise) gives the equations for a and b.