how to calculate shortest 2D distance between a point and a line segment in all cases in C, C# / .NET 2.0 or Java? [duplicate]

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Shortest distance between a point and a line segment

i am looking for a way to calculate the minimum distance in all cases. the problems with solutions i found are:

1. Solutions with graphical conceptual drawings show point always on perpendicular from line segment so it's "between line segment's end points". My geometry skills are horrible so i can't verify that these solutions work in all cases.

2. Algorithm solutions are a: with fortran or some other language i don't fully understand, b: are flagged as incomplete by people, c: calling methods/functions that are not described in any way (considered 开发者_StackOverflowtrivial).

Good example of 2 a, b and c is

Shortest distance between a point and a line segment

i have the 2D line segment as double-type co-ordinate pair (x1, y1), (x2,y2) and point as double type co-ordinate (x3,y3). C#/Java/C solutions are all appreciated.

Thanks for your answers & BR: Matti

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Answered also Shortest distance between a point and a line segment because that gathers solutions in all languages. Answer put also here because this questions asks specifically a C# solution. This is modified from http://www.topcoder.com/tc?d1=tutorials&d2=geometry1&module=Static :

//Compute the dot product AB . BC
private double DotProduct(double[] pointA, double[] pointB, double[] pointC)
{
    double[] AB = new double[2];
    double[] BC = new double[2];
    AB[0] = pointB[0] - pointA[0];
    AB[1] = pointB[1] - pointA[1];
    BC[0] = pointC[0] - pointB[0];
    BC[1] = pointC[1] - pointB[1];
    double dot = AB[0] * BC[0] + AB[1] * BC[1];

    return dot;
}

//Compute the cross product AB x AC
private double CrossProduct(double[] pointA, double[] pointB, double[] pointC)
{
    double[] AB = new double[2];
    double[] AC = new double[2];
    AB[0] = pointB[0] - pointA[0];
    AB[1] = pointB[1] - pointA[1];
    AC[0] = pointC[0] - pointA[0];
    AC[1] = pointC[1] - pointA[1];
    double cross = AB[0] * AC[1] - AB[1] * AC[0];

    return cross;
}

//Compute the distance from A to B
double Distance(double[] pointA, double[] pointB)
{
    double d1 = pointA[0] - pointB[0];
    double d2 = pointA[1] - pointB[1];

    return Math.Sqrt(d1 * d1 + d2 * d2);
}

//Compute the distance from AB to C
//if isSegment is true, AB is a segment, not a line.
double LineToPointDistance2D(double[] pointA, double[] pointB, double[] pointC, 
    bool isSegment)
{
    double dist = CrossProduct(pointA, pointB, pointC) / Distance(pointA, pointB);
    if (isSegment)
    {
        double dot1 = DotProduct(pointA, pointB, pointC);
        if (dot1 > 0) 
            return Distance(pointB, pointC);

        double dot2 = DotProduct(pointB, pointA, pointC);
        if (dot2 > 0) 
            return Distance(pointA, pointC);
    }
    return Math.Abs(dist);
} 

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If you have line

L: A * x + B * y + C = 0

Then distance from this line to point (x1, y1) is abs(A * x1 + B * y1 + C) / sqrt(A * A + B * B). in your case if you has interval, (xa, ya); (xb, yb) you should find min( distance(x1, y1, xa, ya), distance(x1, y1, xb, yb)) then see if perpendecular from (x1, y1) to line L is on the interval, then the answer is the distance is it. otherwise min of two distances.