Hi,

i would like to know how to check as simple as possible if a point lies on a line segment (which is finite). My line has a start (x_s,y_s) and end (x_e,y_e) point and i want to check if point (x_p,y_p) is on the line, how do i do that? The only source code i found does only check if the point is on the line but not on the segment...Any help is very much appreciated.

Thanks for the rectangle-hint! I did it like this and i think it is working now:

        Line.prototype.isOnLine = function (pos) {
            var lowerXBound = Math.min(this.start.x, this.end.x);
            var upperXBound = Math.max(this.start.x, this.end.x);
            var lowerYBound = Math.min(this.start.y, this.end.y);
            var upperYBound = Math.max(this.start.y, this.end.y);
            return pos.x >= lowerXBound && pos.x <= upperXBound && pos.y >= lowerYBound && pos.y <= upperYBound;
        }

Have a nice day!

First get the closest point p' on the infinite line going through s and e and let d = |p' - s|. We can calculate d with vector projection: d = ((e - s) · (p - s)) / |(e - s)|

Knowing d, we can easily get p' as follows: p' = s + d * ((e - s) / |(e - s)|)

Now we can check if p lies inside the rectangle as mentioned earlier by Álvaro. If ε < d < |(e - s)| + ε and |p - p'| < ε your point lies inside the rectangle.

You have point A(x_s,y_s) and point B(x_e,y_e) and want to check if point C(x_p,y_p) is on the line segment L<x_e - x_s, y_e - y_s>.
You can check if vectors AB and AC are colinear. If their cross-product is a zero vector then they are and C is on L. That would be a fast check. If so now you have to check if it is in the bounds - you do this by using the dot product. If 0 < \(\overrightarrow{AB}\bullet\overrightarrow{AC}\) < \(\overrightarrow{AB}\bullet\overrightarrow{AB}\) then C in on L inbetween A and B

Here's all the math for a problem like this: http://www.randygaul.net/2014/07/23/distance-point-to-line-segment/ After getting a squared distance make sure the distance value is below some threshold, and consider below the threshold as "on the line segment".