Find if point lays on line segment

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抹茶落季
抹茶落季 2020-11-30 04:01

I have line segment defined by two points A(x1,y1,z1) and B(x2,y2,z2) and point p(x,y,z). How can I check if the point lays on the line segment?

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  • 2020-11-30 04:36

    Or let the dotnet do the heavy lifting for you if using visual studio use a GraphicsPath

    this will also allow you to add tolerances for if just clicked outside the line.

    using (Drawing2D.GraphicsPath gp = new Drawing2D.GraphicsPath())
    {
        gp.AddLine(new Point(x1, y1), new Point(x2, y2));
    
        // Make the line as wide as needed (make this larger to allow clicking slightly outside the line) 
        using (Pen objPen = new Pen(Color.Black, 6))
        {
            gp.Widen(objPen);
        }
    
        if (gp.IsVisible(Mouse.x, Mouse.y))
        {
            // The line was clicked
        }
    }
    
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  • 2020-11-30 04:39

    Here's some C# code for the 2D case:

    public static bool PointOnLineSegment(PointD pt1, PointD pt2, PointD pt, double epsilon = 0.001)
    {
      if (pt.X - Math.Max(pt1.X, pt2.X) > epsilon || 
          Math.Min(pt1.X, pt2.X) - pt.X > epsilon || 
          pt.Y - Math.Max(pt1.Y, pt2.Y) > epsilon || 
          Math.Min(pt1.Y, pt2.Y) - pt.Y > epsilon)
        return false;
    
      if (Math.Abs(pt2.X - pt1.X) < epsilon)
        return Math.Abs(pt1.X - pt.X) < epsilon || Math.Abs(pt2.X - pt.X) < epsilon;
      if (Math.Abs(pt2.Y - pt1.Y) < epsilon)
        return Math.Abs(pt1.Y - pt.Y) < epsilon || Math.Abs(pt2.Y - pt.Y) < epsilon;
    
      double x = pt1.X + (pt.Y - pt1.Y) * (pt2.X - pt1.X) / (pt2.Y - pt1.Y);
      double y = pt1.Y + (pt.X - pt1.X) * (pt2.Y - pt1.Y) / (pt2.X - pt1.X);
    
      return Math.Abs(pt.X - x) < epsilon || Math.Abs(pt.Y - y) < epsilon;
    }
    
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  • 2020-11-30 04:40

    Your segment is best defined by parametric equation

    for all points on your segment, following equation holds: x = x1 + (x2 - x1) * p y = y1 + (y2 - y1) * p z = z1 + (z2 - z1) * p

    Where p is a number in [0;1]

    So, if there is a p such that your point coordinates satisfy those 3 equations, your point is on this line. And it p is between 0 and 1 - it is also on line segment

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  • 2020-11-30 04:43

    in case if someone looks for inline version:

    public static bool PointOnLine2D (this Vector2 p, Vector2 a, Vector2 b, float t = 1E-03f)
    {
        // ensure points are collinear
        var zero = (b.x - a.x) * (p.y - a.y) - (p.x - a.x) * (b.y - a.y);
        if (zero > t || zero < -t) return false;
    
        // check if x-coordinates are not equal
        if (a.x - b.x > t || b.x - a.x > t)
            // ensure x is between a.x & b.x (use tolerance)
            return a.x > b.x
                ? p.x + t > b.x && p.x - t < a.x
                : p.x + t > a.x && p.x - t < b.x;
    
        // ensure y is between a.y & b.y (use tolerance)
        return a.y > b.y
            ? p.y + t > b.y && p.y - t < a.y
            : p.y + t > a.y && p.y - t < b.y;
    }
    
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  • 2020-11-30 04:43

    I use this to calculate the distance AB between points a and b.

    static void Main(string[] args)
    {
            double AB = segment(0, 1, 0, 4);
            Console.WriteLine("Length of segment AB: {0}",AB);
    }
    
    static double segment (int ax,int ay, int bx, int by)
    {
        Vector a = new Vector(ax,ay);
        Vector b = new Vector(bx,by);
        Vector c = (a & b);
        return Math.Sqrt(c.X + c.Y);
    }
    
    struct Vector
    {
        public readonly float X;
        public readonly float Y;
    
        public Vector(float x, float y)
        {
            this.X = x;
            this.Y = y;
        }
        public static Vector operator &(Vector a, Vector b)  
        {
            return new Vector((b.X - a.X) * (b.X - a.X), (b.Y - a.Y) * (b.Y - a.Y));
        }
    }
    

    based on Calculate a point along the line A-B at a given distance from A

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  • 2020-11-30 04:44

    Based on Konstantin's answer above, here is some C code to find if a point is actually on a FINITE line segment. This takes into account horizontal/vertical line segments. This also takes in to account that floating point numbers are never really "exact" when comparing them with one another. The default epsilon of 0.001f will suffice in most cases. This is for 2D lines... adding "Z" would be trivial. PointF class is from GDI+, which is basically just: struct PointF{float X,Y};

    Hope this helps!

    #define DEFFLEQEPSILON 0.001
    #define FLOAT_EQE(x,v,e)((((v)-(e))<(x))&&((x)<((v)+(e))))
    
    static bool Within(float fl, float flLow, float flHi, float flEp=DEFFLEQEPSILON){
        if((fl>flLow) && (fl<flHi)){ return true; }
        if(FLOAT_EQE(fl,flLow,flEp) || FLOAT_EQE(fl,flHi,flEp)){ return true; }
        return false;
    }
    
    static bool PointOnLine(const PointF& ptL1, const PointF& ptL2, const PointF& ptTest, float flEp=DEFFLEQEPSILON){
        bool bTestX = true;
        const float flX = ptL2.X-ptL1.X;
        if(FLOAT_EQE(flX,0.0f,flEp)){
            // vertical line -- ptTest.X must equal ptL1.X to continue
            if(!FLOAT_EQE(ptTest.X,ptL1.X,flEp)){ return false; }
            bTestX = false;
        }
        bool bTestY = true;
        const float flY = ptL2.Y-ptL1.Y;
        if(FLOAT_EQE(flY,0.0f,flEp)){
            // horizontal line -- ptTest.Y must equal ptL1.Y to continue
            if(!FLOAT_EQE(ptTest.Y,ptL1.Y,flEp)){ return false; }
            bTestY = false;
        }
        // found here: http://stackoverflow.com/a/7050309
        // x = x1 + (x2 - x1) * p
        // y = y1 + (y2 - y1) * p
        // solve for p:
        const float pX = bTestX?((ptTest.X-ptL1.X)/flX):0.5f;
        const float pY = bTestY?((ptTest.Y-ptL1.Y)/flY):0.5f;
        return Within(pX,0.0f,1.0f,flEp) && Within(pY,0.0f,1.0f,flEp);
    }
    
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