Question

我在C＃WPF中有两个Line对象，我正在尝试构造一个方法来计算出线相交的坐标（如果有的话）。在让自己头疼提醒自己高中数学这样做之后，我无法弄清楚如何将其映射到编程格式 - 任何人都知道如何做到这一点？

非常感谢，贝基

Answer 1

我想你的线对象是由两点组成的。你应该做的是得到两条线的方程式。

然后你应该解决以下等式：

equation-line1 = equation-line2

计算线的斜率：

    float _slope = 1e+10;
    if ((p2.x - p1.x) != 0)
        _slope = (p2.y - p1.y) / (p2.x - p1.x);

p1 =你的第1行 p2 =你的第2行

一条线的方程式：

y = ax + b

a =您计算的斜率 b =相交

求解等式：

    a1 = p1.y - b1 * p1.x;
    a2 = q1.y - b2 * q1.x;

    x = (a1 - a2) / (b2 - b1);
    y = a2 + b2 * x;

乏：

b1 =斜率线1
b2 = slop line 2
q1 =第二行的第1点

因此，x和y是两条线相交的点的坐标

Answer 2

这更像是一个真实的答案，而不是一个实际的答案，因为如果你所做的只是相交两行，那就非常慢。但我认为值得一提。

WPF能够交叉任意两个形状轮廓，包括两条线，并告诉你交叉点的位置。

以下是交叉两个轮廓（边缘，而不是填充区域）的一般技术：

var shape1 = ...;
var shape2 = ...;

var thinPen = new Pen(Brush.Transparent, 0.001);

var geometry1 = shape1.RenderedGeometry.GetWidenedPathGeometry(thinPen);
var geometry2 = shape2.RenderedGeometry.GetWidenedPathGeometry(thinPen);

var combined = Geometry.Combine(
                 geometry1,
                 geometry2,
                 GeometryCombineMode.Intersect,
                 shape2.TransformToVisual(shape1));

var bounds = combined.GetRenderBounds(thinPen);

如果已知形状具有相同的位置，则shape2.TransformToVisual(shape1)调用中的Geometry.Combine可以替换为null。

这项技术非常有用，例如，如果你需要将一条直线与任意曲线相交。

Answer 3

如何找到两条线/线段的交叉点/带有矩形的光线

public class LineEquation{
    public LineEquation(Point start, Point end){
        Start = start;
        End = end;

        IsVertical = Math.Abs(End.X - start.X) < 0.00001f;
        M = (End.Y - Start.Y)/(End.X - Start.X);
        A = -M;
        B = 1;
        C = Start.Y - M*Start.X;
    }

    public bool IsVertical { get; private set; }

    public double M { get; private set; }

    public Point Start { get; private set; }
    public Point End { get; private set; }

    public double A { get; private set; }
    public double B { get; private set; }
    public double C { get; private set; }

    public bool IntersectsWithLine(LineEquation otherLine, out Point intersectionPoint){
        intersectionPoint = new Point(0, 0);
        if (IsVertical && otherLine.IsVertical)
            return false;
        if (IsVertical || otherLine.IsVertical){
            intersectionPoint = GetIntersectionPointIfOneIsVertical(otherLine, this);
            return true;
        }
        double delta = A*otherLine.B - otherLine.A*B;
        bool hasIntersection = Math.Abs(delta - 0) > 0.0001f;
        if (hasIntersection){
            double x = (otherLine.B*C - B*otherLine.C)/delta;
            double y = (A*otherLine.C - otherLine.A*C)/delta;
            intersectionPoint = new Point(x, y);
        }
        return hasIntersection;
    }

    private static Point GetIntersectionPointIfOneIsVertical(LineEquation line1, LineEquation line2){
        LineEquation verticalLine = line2.IsVertical ? line2 : line1;
        LineEquation nonVerticalLine = line2.IsVertical ? line1 : line2;

        double y = (verticalLine.Start.X - nonVerticalLine.Start.X)*
                   (nonVerticalLine.End.Y - nonVerticalLine.Start.Y)/
                   ((nonVerticalLine.End.X - nonVerticalLine.Start.X)) +
                   nonVerticalLine.Start.Y;
        double x = line1.IsVertical ? line1.Start.X : line2.Start.X;
        return new Point(x, y);
    }

    public bool IntersectWithSegementOfLine(LineEquation otherLine, out Point intersectionPoint){
        bool hasIntersection = IntersectsWithLine(otherLine, out intersectionPoint);
        if (hasIntersection)
            return intersectionPoint.IsBetweenTwoPoints(otherLine.Start, otherLine.End);
        return false;
    }

    public bool GetIntersectionLineForRay(Rect rectangle, out LineEquation intersectionLine){
        if (Start == End){
            intersectionLine = null;
            return false;
        }
        IEnumerable<LineEquation> lines = rectangle.GetLinesForRectangle();
        intersectionLine = new LineEquation(new Point(0, 0), new Point(0, 0));
        var intersections = new Dictionary<LineEquation, Point>();
        foreach (LineEquation equation in lines){
            Point point;
            if (IntersectWithSegementOfLine(equation, out point))
                intersections[equation] = point;
        }
        if (!intersections.Any())
            return false;

        var intersectionPoints = new SortedDictionary<double, Point>();
        foreach (var intersection in intersections){
            if (End.IsBetweenTwoPoints(Start, intersection.Value) ||
                intersection.Value.IsBetweenTwoPoints(Start, End)){
                double distanceToPoint = Start.DistanceToPoint(intersection.Value);
                intersectionPoints[distanceToPoint] = intersection.Value;
            }
        }
        if (intersectionPoints.Count == 1){
            Point endPoint = intersectionPoints.First().Value;
            intersectionLine = new LineEquation(Start, endPoint);
            return true;
        }

        if (intersectionPoints.Count == 2){
            Point start = intersectionPoints.First().Value;
            Point end = intersectionPoints.Last().Value;
            intersectionLine = new LineEquation(start, end);
            return true;
        }

        return false;
    }

    public override string ToString(){
        return "[" + Start + "], [" + End + "]";
    }
}

完整样本描述为here

WPF两个Line对象的交点坐标

3 个答案: