许多答案把所有的计算都打包成一个函数。如果您需要计算直线斜率、y轴截距或x轴截距，以便在代码的其他地方使用，那么这些计算将是冗余的。我分离出了各自的函数，使用了明显的变量名，并注释了我的代码以使其更易于理解。我需要知道直线是否无限超出它们的端点，所以在JavaScript中:

http://jsfiddle.net/skibulk/evmqq00u/

var point_a = {x:0, y:10},
    point_b = {x:12, y:12},
    point_c = {x:10, y:0},
    point_d = {x:0, y:0},
    slope_ab = slope(point_a, point_b),
    slope_bc = slope(point_b, point_c),
    slope_cd = slope(point_c, point_d),
    slope_da = slope(point_d, point_a),
    yint_ab = y_intercept(point_a, slope_ab),
    yint_bc = y_intercept(point_b, slope_bc),
    yint_cd = y_intercept(point_c, slope_cd),
    yint_da = y_intercept(point_d, slope_da),
    xint_ab = x_intercept(point_a, slope_ab, yint_ab),
    xint_bc = x_intercept(point_b, slope_bc, yint_bc),
    xint_cd = x_intercept(point_c, slope_cd, yint_cd),
    xint_da = x_intercept(point_d, slope_da, yint_da),
    point_aa = intersect(slope_da, yint_da, xint_da, slope_ab, yint_ab, xint_ab),
    point_bb = intersect(slope_ab, yint_ab, xint_ab, slope_bc, yint_bc, xint_bc),
    point_cc = intersect(slope_bc, yint_bc, xint_bc, slope_cd, yint_cd, xint_cd),
    point_dd = intersect(slope_cd, yint_cd, xint_cd, slope_da, yint_da, xint_da);

console.log(point_a, point_b, point_c, point_d);
console.log(slope_ab, slope_bc, slope_cd, slope_da);
console.log(yint_ab, yint_bc, yint_cd, yint_da);
console.log(xint_ab, xint_bc, xint_cd, xint_da);
console.log(point_aa, point_bb, point_cc, point_dd);

function slope(point_a, point_b) {
  var i = (point_b.y - point_a.y) / (point_b.x - point_a.x);
  if (i === -Infinity) return Infinity;
  if (i === -0) return 0;
  return i;
}

function y_intercept(point, slope) {
    // Horizontal Line
    if (slope == 0) return point.y;
  // Vertical Line
    if (slope == Infinity)
  {
    // THE Y-Axis
    if (point.x == 0) return Infinity;
    // No Intercept
    return null;
  }
  // Angled Line
  return point.y - (slope * point.x);
}

function x_intercept(point, slope, yint) {
    // Vertical Line
    if (slope == Infinity) return point.x;
  // Horizontal Line
    if (slope == 0)
  {
    // THE X-Axis
    if (point.y == 0) return Infinity;
    // No Intercept
    return null;
  }
  // Angled Line
  return -yint / slope;
}

// Intersection of two infinite lines
function intersect(slope_a, yint_a, xint_a, slope_b, yint_b, xint_b) {
  if (slope_a == slope_b)
  {
    // Equal Lines
    if (yint_a == yint_b && xint_a == xint_b) return Infinity;
    // Parallel Lines
    return null;
  }
  // First Line Vertical
    if (slope_a == Infinity)
  {
    return {
        x: xint_a,
      y: (slope_b * xint_a) + yint_b
    };
  }
  // Second Line Vertical
    if (slope_b == Infinity)
  {
    return {
        x: xint_b,
      y: (slope_a * xint_b) + yint_a
    };
  }
  // Not Equal, Not Parallel, Not Vertical
  var i = (yint_b - yint_a) / (slope_a - slope_b);
  return {
    x: i,
    y: (slope_a * i) + yint_a
  };
}

人们似乎对Gavin的答案很感兴趣，cortijon在评论中提出了一个javascript版本，iMalc提供了一个计算量略少的版本。一些人指出了各种代码建议的缺点，另一些人则评论了一些代码建议的效率。

iMalc通过Gavin的答案提供的算法是我目前在一个javascript项目中使用的算法，我只是想在这里提供一个清理过的版本，如果它可以帮助到任何人的话。

// Some variables for reuse, others may do this differently
var p0x, p1x, p2x, p3x, ix,
    p0y, p1y, p2y, p3y, iy,
    collisionDetected;

// do stuff, call other functions, set endpoints...

// note: for my purpose I use |t| < |d| as opposed to
// |t| <= |d| which is equivalent to 0 <= t < 1 rather than
// 0 <= t <= 1 as in Gavin's answer - results may vary

var lineSegmentIntersection = function(){
    var d, dx1, dx2, dx3, dy1, dy2, dy3, s, t;

    dx1 = p1x - p0x;      dy1 = p1y - p0y;
    dx2 = p3x - p2x;      dy2 = p3y - p2y;
    dx3 = p0x - p2x;      dy3 = p0y - p2y;

    collisionDetected = 0;

    d = dx1 * dy2 - dx2 * dy1;

    if(d !== 0){
        s = dx1 * dy3 - dx3 * dy1;
        if((s <= 0 && d < 0 && s >= d) || (s >= 0 && d > 0 && s <= d)){
            t = dx2 * dy3 - dx3 * dy2;
            if((t <= 0 && d < 0 && t > d) || (t >= 0 && d > 0 && t < d)){
                t = t / d;
                collisionDetected = 1;
                ix = p0x + t * dx1;
                iy = p0y + t * dy1;
            }
        }
    }
};

我试过其中一些答案，但它们对我不起作用(对不起伙计们);在网上搜索之后，我找到了这个。

对他的代码做了一点修改，我现在有了这个函数，它将返回交点，如果没有找到交点，它将返回- 1,1。

    Public Function intercetion(ByVal ax As Integer, ByVal ay As Integer, ByVal bx As Integer, ByVal by As Integer, ByVal cx As Integer, ByVal cy As Integer, ByVal dx As Integer, ByVal dy As Integer) As Point
    '//  Determines the intersection point of the line segment defined by points A and B
    '//  with the line segment defined by points C and D.
    '//
    '//  Returns YES if the intersection point was found, and stores that point in X,Y.
    '//  Returns NO if there is no determinable intersection point, in which case X,Y will
    '//  be unmodified.

    Dim distAB, theCos, theSin, newX, ABpos As Double

    '//  Fail if either line segment is zero-length.
    If ax = bx And ay = by Or cx = dx And cy = dy Then Return New Point(-1, -1)

    '//  Fail if the segments share an end-point.
    If ax = cx And ay = cy Or bx = cx And by = cy Or ax = dx And ay = dy Or bx = dx And by = dy Then Return New Point(-1, -1)

    '//  (1) Translate the system so that point A is on the origin.
    bx -= ax
    by -= ay
    cx -= ax
    cy -= ay
    dx -= ax
    dy -= ay

    '//  Discover the length of segment A-B.
    distAB = Math.Sqrt(bx * bx + by * by)

    '//  (2) Rotate the system so that point B is on the positive X axis.
    theCos = bx / distAB
    theSin = by / distAB
    newX = cx * theCos + cy * theSin
    cy = cy * theCos - cx * theSin
    cx = newX
    newX = dx * theCos + dy * theSin
    dy = dy * theCos - dx * theSin
    dx = newX

    '//  Fail if segment C-D doesn't cross line A-B.
    If cy < 0 And dy < 0 Or cy >= 0 And dy >= 0 Then Return New Point(-1, -1)

    '//  (3) Discover the position of the intersection point along line A-B.
    ABpos = dx + (cx - dx) * dy / (dy - cy)

    '//  Fail if segment C-D crosses line A-B outside of segment A-B.
    If ABpos < 0 Or ABpos > distAB Then Return New Point(-1, -1)

    '//  (4) Apply the discovered position to line A-B in the original coordinate system.
    '*X=Ax+ABpos*theCos
    '*Y=Ay+ABpos*theSin

    '//  Success.
    Return New Point(ax + ABpos * theCos, ay + ABpos * theSin)
End Function

问题可以简化成这样一个问题:从A到B和从C到D的两条直线相交吗?然后你可以问它四次(在直线和矩形的四条边之间)。

这是做这个的矢量数学。假设A到B的直线就是问题中的直线C到D的直线是其中一条矩形直线。我的表示法是Ax是A的x坐标Cy是c的y坐标“*”表示点积，例如A*B = Ax*Bx + Ay*By。

E = B-A = ( Bx-Ax, By-Ay )
F = D-C = ( Dx-Cx, Dy-Cy ) 
P = ( -Ey, Ex )
h = ( (A-C) * P ) / ( F * P )

h是键。如果h在0和1之间，两条线相交，否则不相交。如果F*P为零，当然不能进行计算，但在这种情况下，直线是平行的，因此只有在明显的情况下才相交。

交点是C + F*h。

更多的乐趣:

如果h恰好等于0或1，两条直线的端点相交。你可以认为这是一个“交集”，也可以认为不是。

具体来说，h是直线长度乘以多少才能恰好与另一条直线相交。

因此，如果h<0，这意味着矩形线在给定直线的“后面”(“方向”是“从A到B”)，如果h>1，矩形线在给定直线的“前面”。

推导:

A和C是指向直线起点的向量;E和F是由A和C端点组成的直线。

对于平面上任意两条不平行线，必须恰好有一对标量g和h，使得这个方程成立:

A + E*g = C + F*h

为什么?因为两条不平行线必须相交，这意味着你可以将这两条线按一定比例缩放并相互接触。

(起初，这看起来像一个有两个未知数的方程!但当你考虑到这是一个二维矢量方程时，它就不是，这意味着这是一对x和y的方程)

我们必须消去其中一个变量。一个简单的方法是使E项为零。要做到这一点，用一个向量对方程两边做点积这个向量与E点乘到0，我把上面的向量称为P，我做了E的明显变换。

你现在有:

A*P = C*P + F*P*h
(A-C)*P = (F*P)*h
( (A-C)*P ) / (F*P) = h

找到两条线段的正确交点是一项具有大量边缘情况的非简单任务。下面是一个用Java编写的、有效的、经过测试的解决方案。

本质上，在求两条线段的交点时，有三种情况会发生:

线段不相交 有一个唯一的交点 交点是另一段

注意:在代码中，我假设x1 = x2和y1 = y2的线段(x1, y1)， (x2, y2)是有效的线段。从数学上讲，线段由不同的点组成，但为了完整起见，我在这个实现中允许线段作为点。

代码是从我的github回购

/**
 * This snippet finds the intersection of two line segments.
 * The intersection may either be empty, a single point or the
 * intersection is a subsegment there's an overlap.
 */

import static java.lang.Math.abs;
import static java.lang.Math.max;
import static java.lang.Math.min;

import java.util.ArrayList;
import java.util.List;

public class LineSegmentLineSegmentIntersection {

  // Small epsilon used for double value comparison.
  private static final double EPS = 1e-5;

  // 2D Point class.
  public static class Pt {
    double x, y;
    public Pt(double x, double y) {
      this.x = x; 
      this.y = y;
    }
    public boolean equals(Pt pt) {
      return abs(x - pt.x) < EPS && abs(y - pt.y) < EPS;
    }
  }

  // Finds the orientation of point 'c' relative to the line segment (a, b)
  // Returns  0 if all three points are collinear.
  // Returns -1 if 'c' is clockwise to segment (a, b), i.e right of line formed by the segment.
  // Returns +1 if 'c' is counter clockwise to segment (a, b), i.e left of line
  // formed by the segment.
  public static int orientation(Pt a, Pt b, Pt c) {
    double value = (b.y - a.y) * (c.x - b.x) - 
                   (b.x - a.x) * (c.y - b.y);
    if (abs(value) < EPS) return 0;
    return (value > 0) ? -1 : +1;
  }

  // Tests whether point 'c' is on the line segment (a, b).
  // Ensure first that point c is collinear to segment (a, b) and
  // then check whether c is within the rectangle formed by (a, b)
  public static boolean pointOnLine(Pt a, Pt b, Pt c) {
    return orientation(a, b, c) == 0 && 
           min(a.x, b.x) <= c.x && c.x <= max(a.x, b.x) && 
           min(a.y, b.y) <= c.y && c.y <= max(a.y, b.y);
  }

  // Determines whether two segments intersect.
  public static boolean segmentsIntersect(Pt p1, Pt p2, Pt p3, Pt p4) {

    // Get the orientation of points p3 and p4 in relation
    // to the line segment (p1, p2)
    int o1 = orientation(p1, p2, p3);
    int o2 = orientation(p1, p2, p4);
    int o3 = orientation(p3, p4, p1);
    int o4 = orientation(p3, p4, p2);

    // If the points p1, p2 are on opposite sides of the infinite
    // line formed by (p3, p4) and conversly p3, p4 are on opposite
    // sides of the infinite line formed by (p1, p2) then there is
    // an intersection.
    if (o1 != o2 && o3 != o4) return true;

    // Collinear special cases (perhaps these if checks can be simplified?)
    if (o1 == 0 && pointOnLine(p1, p2, p3)) return true;
    if (o2 == 0 && pointOnLine(p1, p2, p4)) return true;
    if (o3 == 0 && pointOnLine(p3, p4, p1)) return true;
    if (o4 == 0 && pointOnLine(p3, p4, p2)) return true;

    return false;
  }

  public static List<Pt> getCommonEndpoints(Pt p1, Pt p2, Pt p3, Pt p4) {

    List<Pt> points = new ArrayList<>();

    if (p1.equals(p3)) {
      points.add(p1);
      if (p2.equals(p4)) points.add(p2);

    } else if (p1.equals(p4)) {
      points.add(p1);
      if (p2.equals(p3)) points.add(p2);

    } else if (p2.equals(p3)) {
      points.add(p2);
      if (p1.equals(p4)) points.add(p1);

    } else if (p2.equals(p4)) {
      points.add(p2);
      if (p1.equals(p3)) points.add(p1);
    }

    return points;
  }

  // Finds the intersection point(s) of two line segments. Unlike regular line 
  // segments, segments which are points (x1 = x2 and y1 = y2) are allowed.
  public static Pt[] lineSegmentLineSegmentIntersection(Pt p1, Pt p2, Pt p3, Pt p4) {

    // No intersection.
    if (!segmentsIntersect(p1, p2, p3, p4)) return new Pt[]{};

    // Both segments are a single point.
    if (p1.equals(p2) && p2.equals(p3) && p3.equals(p4))
      return new Pt[]{p1};

    List<Pt> endpoints = getCommonEndpoints(p1, p2, p3, p4);
    int n = endpoints.size();

    // One of the line segments is an intersecting single point.
    // NOTE: checking only n == 1 is insufficient to return early
    // because the solution might be a sub segment.
    boolean singleton = p1.equals(p2) || p3.equals(p4);
    if (n == 1 && singleton) return new Pt[]{endpoints.get(0)};

    // Segments are equal.
    if (n == 2) return new Pt[]{endpoints.get(0), endpoints.get(1)};

    boolean collinearSegments = (orientation(p1, p2, p3) == 0) && 
                                (orientation(p1, p2, p4) == 0);

    // The intersection will be a sub-segment of the two
    // segments since they overlap each other.
    if (collinearSegments) {

      // Segment #2 is enclosed in segment #1
      if (pointOnLine(p1, p2, p3) && pointOnLine(p1, p2, p4))
        return new Pt[]{p3, p4};

      // Segment #1 is enclosed in segment #2
      if (pointOnLine(p3, p4, p1) && pointOnLine(p3, p4, p2))
        return new Pt[]{p1, p2};

      // The subsegment is part of segment #1 and part of segment #2.
      // Find the middle points which correspond to this segment.
      Pt midPoint1 = pointOnLine(p1, p2, p3) ? p3 : p4;
      Pt midPoint2 = pointOnLine(p3, p4, p1) ? p1 : p2;

      // There is actually only one middle point!
      if (midPoint1.equals(midPoint2)) return new Pt[]{midPoint1};

      return new Pt[]{midPoint1, midPoint2};
    }

    /* Beyond this point there is a unique intersection point. */

    // Segment #1 is a vertical line.
    if (abs(p1.x - p2.x) < EPS) {
      double m = (p4.y - p3.y) / (p4.x - p3.x);
      double b = p3.y - m * p3.x;
      return new Pt[]{new Pt(p1.x, m * p1.x + b)};
    }

    // Segment #2 is a vertical line.
    if (abs(p3.x - p4.x) < EPS) {
      double m = (p2.y - p1.y) / (p2.x - p1.x);
      double b = p1.y - m * p1.x;
      return new Pt[]{new Pt(p3.x, m * p3.x + b)};
    }

    double m1 = (p2.y - p1.y) / (p2.x - p1.x);
    double m2 = (p4.y - p3.y) / (p4.x - p3.x);
    double b1 = p1.y - m1 * p1.x;
    double b2 = p3.y - m2 * p3.x;
    double x = (b2 - b1) / (m1 - m2);
    double y = (m1 * b2 - m2 * b1) / (m1 - m2);

    return new Pt[]{new Pt(x, y)};
  }

}

下面是一个简单的用法示例:

  public static void main(String[] args) {

    // Segment #1 is (p1, p2), segment #2 is (p3, p4)
    Pt p1, p2, p3, p4;

    p1 = new Pt(-2, 4); p2 = new Pt(3, 3);
    p3 = new Pt(0, 0);  p4 = new Pt(2, 4);
    Pt[] points = lineSegmentLineSegmentIntersection(p1, p2, p3, p4);
    Pt point = points[0];

    // Prints: (1.636, 3.273)
    System.out.printf("(%.3f, %.3f)\n", point.x, point.y);

    p1 = new Pt(-10, 0); p2 = new Pt(+10, 0);
    p3 = new Pt(-5, 0);  p4 = new Pt(+5, 0);
    points = lineSegmentLineSegmentIntersection(p1, p2, p3, p4);
    Pt point1 = points[0], point2 = points[1];

    // Prints: (-5.000, 0.000) (5.000, 0.000)
    System.out.printf("(%.3f, %.3f) (%.3f, %.3f)\n", point1.x, point1.y, point2.x, point2.y);
  }

只是想提一下，一个很好的解释和明确的解决方案可以在数字食谱系列中找到。我有这本书的第三版，答案在1117页21.4节。另一种不同命名的解决方案可以在玛丽娜·加夫里洛娃(Marina Gavrilova)的论文中找到。在我看来，她的解决办法要简单一些。

我的实现如下:

bool NuGeometry::IsBetween(const double& x0, const double& x, const double& x1){
   return (x >= x0) && (x <= x1);
}

bool NuGeometry::FindIntersection(const double& x0, const double& y0, 
     const double& x1, const double& y1,
     const double& a0, const double& b0, 
     const double& a1, const double& b1, 
     double& xy, double& ab) {
   // four endpoints are x0, y0 & x1,y1 & a0,b0 & a1,b1
   // returned values xy and ab are the fractional distance along xy and ab
   // and are only defined when the result is true

   bool partial = false;
   double denom = (b0 - b1) * (x0 - x1) - (y0 - y1) * (a0 - a1);
   if (denom == 0) {
      xy = -1;
      ab = -1;
   } else {
      xy = (a0 * (y1 - b1) + a1 * (b0 - y1) + x1 * (b1 - b0)) / denom;
      partial = NuGeometry::IsBetween(0, xy, 1);
      if (partial) {
         // no point calculating this unless xy is between 0 & 1
         ab = (y1 * (x0 - a1) + b1 * (x1 - x0) + y0 * (a1 - x1)) / denom; 
      }
   }
   if ( partial && NuGeometry::IsBetween(0, ab, 1)) {
      ab = 1-ab;
      xy = 1-xy;
      return true;
   }  else return false;
}