Line to line intersection point

Question

If i have a line with end points ( x1 , y1 ) ( x2 , y2 ) and
another line that crosses it, and its end points are ( x3 , y3 ) ( x4 , y4 )
how do i get the intersection point ( IPx , IPy ) ?

Answer 1

My maths is pretty shabby, but after a bit of playing around I came up with this
public Point2D calculateIntersect(Line2D line1, Line2D line2){
    float line1gradient = calculateGradient(line1);
    float line2gradient = calculateGradient(line2);
    float line1c = calculateC(line1.getP1(), line1gradient);
    float line2c = calculateC(line2.getP1(), line2gradient);
    float x = (line2c - line1c) / (line1gradient - line2gradient);
    float y = line1gradient * x + line1c;
    return new Point2D.Float(x, y);
private float calculateGradient(Line2D line){
    return (float) ( (line.getP2().getY() - line.getP1().getY()) / (line.getP2().getX() - line.getP1().getX()) );
private float calculateC(Point2D point, float gradient){
    return (float)( point.getY() - ( point.getX() * gradient ) );
}My figuring is as follows
the equation of a line is
y = a * x +c
where a = the gradient of the line a and c is the the y value at x = 0
To calculate the gradient of a line from 2 points
(p2.y - p1.y) / (p2.x - p1.x)
and to calculate the value of c for the line
p.y - p.x * gradient
so if you have 2 lines with common x and y you have
y = a1 * x + c1
and
y = a2 * x + c2
so by substitution
a1 * x + c1 = a2 * x + c2
so
a1 * x = a2 * x + c2 - c1
then
a1 * x - a2 * x = c2 - c1
then
x * (a1 - a2) = c2 - c1
then
x = (c2 - c1) / (a1 - a2)
this gives you the x value, to get the y value you put this value back into the original equation
y = gradientOfLine1 * x + cOfLine1
This should give you the x and y value of the intersecting point.
p.s. Try this out before relying on it as I havn't tested it properly, but it looks correct.
Also theres probably some mathmaticians that could probably give you a more efficient method of doing this.
Hope this helps Michael