Efficient maths algorithm to calculate intersections

Question

For a game I am developing I need an algorithm that can calculate intersections. I have solved the problem, but the way I have done it is really nasty and I am hoping someon

Answer 1

public struct PointD
{
    public double X { get; set; }
    public double Y { get; set; }
}

/// 

/// Find the intersection point between two lines.
/// 
/// The intersection point. A PointD structure.
/// The starting point of first line. A PointD structure.
/// The end point of first line. A PointD structure.
/// The starting point of second line. A PointD structure.
/// The end point of second line. A PointD structure.
/// The intersection position at first line.
/// The intersection position at second line.
/// Returns a boolean. True if there is intersection, false otherwise.
/// The formula is taken from comp.graphics.algorithms Frequently Asked Questions.
public static bool LineIntersect(out PointD IntersectPoint, PointD L1StartPoint, PointD L1EndPoint, PointD L2StartPoint, PointD L2EndPoint, out double L1IntersectPos, out double L2IntersectPos) 
{
    IntersectPoint = new PointD();
    double ay_cy, ax_cx, px, py;
    double dx_cx = L2EndPoint.X - L2StartPoint.X,
        dy_cy = L2EndPoint.Y - L2StartPoint.Y,
        bx_ax = L1EndPoint.X - L1StartPoint.X,
        by_ay = L1EndPoint.Y - L1StartPoint.Y;

    double de = (bx_ax) * (dy_cy) - (by_ay) * (dx_cx);
    //double tor = 1.0E-10;     //tolerance


    L1IntersectPos = 0;      L2IntersectPos = 0;
    if (Math.Abs(de)<0.01)
        return false;
    //if (de > -tor && de < tor) return false; //line is in parallel

    ax_cx = L1StartPoint.X - L2StartPoint.X;
    ay_cy = L1StartPoint.Y - L2StartPoint.Y;
    double r = ((ay_cy) * (dx_cx) - (ax_cx) * (dy_cy)) / de;
    double s = ((ay_cy) * (bx_ax) - (ax_cx) * (by_ay)) / de;
    px = L1StartPoint.X + r * (bx_ax);
    py = L1StartPoint.Y + r * (by_ay);

    IntersectPoint.X = px;  //return the intersection point
    IntersectPoint.Y = py;  //return the intersection position
    L1IntersectPos = r;      L2IntersectPos = s;

    return true; //indicate there is intersection
}

To check whether the intersection point is not beyond the original length of the line, just make sure that 0 and 0