如何判断一个点是否属于某一行？

Question

如何判断一个点是否属于某一行？

如果可能的话,可以理解例子.

Answer 1

在最简单的形式中,只需将坐标插入线方程并检查是否相等.

鉴于:

Point p (X=4, Y=5)
Line l (Slope=1, YIntersect=1)

插入X和Y:

   Y = Slope * X + YIntersect
=> 5 = 1 * 4 + 1
=> 5 = 5

所以,是的,重点是在线.

如果您的线条以(X1,Y1),(X2,Y2)形式表示,那么您可以使用以下公式计算坡度:

 Slope = (y1 - y2) / (x1-x2)

然后得到Y-Intersect:

 YIntersect = - Slope * X1 + Y1;

编辑:我修复了Y-Intersect(已经是X1/Y1 ...)

你必须检查x1 - x2不是0.如果是,那么检查点是否在线是一个简单的问题,检查你的点中的Y值是否等于x1或x2.另外,检查点的X不是'x1'或'x2'.

Answer 2

我刚刚写了一个函数来处理一些额外的要求,因为我在绘图应用程序中使用了这个检查:

• 模糊性 - 必须有一些错误的余地,因为该功能用于通过单击来选择线条.
• 该线有一个EndPoint和一个StartPoint,没有无限的线.
• 必须处理直的垂直和水平线,(x2 - x1)== 0导致在其他答案中除以零.

private const double SELECTION_FUZZINESS = 3;

internal override bool ContainsPoint(Point point)
{
    LineGeometry lineGeo = geometry as LineGeometry;
    Point leftPoint;
    Point rightPoint;

    // Normalize start/end to left right to make the offset calc simpler.
    if (lineGeo.StartPoint.X <= lineGeo.EndPoint.X)
    {
        leftPoint   = lineGeo.StartPoint;
        rightPoint  = lineGeo.EndPoint;
    }
    else
    {
        leftPoint   = lineGeo.EndPoint;
        rightPoint  = lineGeo.StartPoint;
    }

    // If point is out of bounds, no need to do further checks.                  
    if (point.X + SELECTION_FUZZINESS < leftPoint.X || rightPoint.X < point.X - SELECTION_FUZZINESS)
        return false;
    else if (point.Y + SELECTION_FUZZINESS < Math.Min(leftPoint.Y, rightPoint.Y) || Math.Max(leftPoint.Y, rightPoint.Y) < point.Y - SELECTION_FUZZINESS)
        return false;

    double deltaX = rightPoint.X - leftPoint.X;
    double deltaY = rightPoint.Y - leftPoint.Y;

    // If the line is straight, the earlier boundary check is enough to determine that the point is on the line.
    // Also prevents division by zero exceptions.
    if (deltaX == 0 || deltaY == 0) 
        return true;

    double slope        = deltaY / deltaX;
    double offset       = leftPoint.Y - leftPoint.X * slope;
    double calculatedY  = point.X * slope + offset;

    // Check calculated Y matches the points Y coord with some easing.
    bool lineContains = point.Y - SELECTION_FUZZINESS <= calculatedY && calculatedY <= point.Y + SELECTION_FUZZINESS;

    return lineContains;            
}

Answer 3

确定点R =(rx,ry)是否位于连接点P =(px,py)和Q =(qx,qy)的线上的最佳方法是检查矩阵的行列式

{{qx - px, qy - py}, {rx - px, ry - py}},

即(qx - px)*(ry - py) - (qy - py)*(rx - px)接近0.此解决方案与其他方案相比具有几个相关的优点:首先,它不需要垂直线的特殊情况,第二,它不分(通常是慢速操作),第三,当线几乎但不是很垂直时,它不会触发不良的浮点行为.

Answer 4

鉴于上线两分L0和L1和测试点P.

               (L1 - L0) * (P - L0)
n = (P - L0) - --------------------- (L1 - L0)
               (L1 - L0) * (L1 - L0)

矢量的范数n是点P与线的距离L0和L1.如果该距离为零或足够小(在舍入误差的情况下),则该点位于该线上.

符号*代表点积.

例

P = (5, 5)

L0 = (0, 10)
L1 = (20, -10)

L1 - L0 = (20, -20)
P  - L0 = (5, -5)

              (20, -20) * (5, -5)
n = (5, -5) - --------------------- (20, -20)
              (20, -20) * (20, -20)

              200
  = (5, -5) - --- (20, -20)
              800

  = (5, -5) - (5, -5)

  = (0, 0)

Answer 5

我认为帕特里克麦克唐纳先生提出了几乎正确的答案,这是对他答案的更正:

public bool IsOnLine(Point endPoint1, Point endPoint2, Point checkPoint)
{
    return (((double)checkPoint.Y - endPoint1.Y)) / ((double)(checkPoint.X - endPoint1.X))
        == ((double)(endPoint2.Y - endPoint1.Y)) / ((double)(endPoint2.X - endPoint1.X));
}

当然还有许多其他正确答案,尤其是Mr.Josh,但我发现这是最好的答案.

感谢evryone.