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
  };
}

曾经在这里被接受的答案是不正确的(它已经被不接受了，所以万岁!)它不能正确地消除所有非交点。简单地说，它可能有效，但也可能失败，特别是在0和1被认为对h有效的情况下。

考虑以下情况:

直线(4,1)-(5,1)和(0,0)-(0,2)

这两条垂线显然不重叠。

= (4,1) B =(5、1) C = (0, 0) D = (0, 2) E = (1) - (4,1) = (1,0) F = (0, 2) - (0, 0) = (0, 2) P = (0, 1) h =((4,1) -(0, 0))点(0,1)/((0,2)点(0,1))= 0

根据上面的答案，这两条线段在端点处相遇(值为0和1)。该端点为:

(0, 0) + (0, 2) * 0 = (0, 0)

So, apparently the two line segments meet at (0,0), which is on line CD, but not on line AB. So what is going wrong? The answer is that the values of 0 and 1 are not valid and only sometimes HAPPEN to correctly predict endpoint intersection. When the extension of one line (but not the other) would meet the line segment, the algorithm predicts an intersection of line segments, but this is not correct. I imagine that by testing starting with AB vs CD and then also testing with CD vs AB, this problem would be eliminated. Only if both fall between 0 and 1 inclusively can they be said to intersect.

如果你必须预测端点，我建议使用向量叉乘法。

-Dan

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;
}

根据t3chb0t的答案:

int intersezione_linee(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4, int& p_x, int& p_y)
{
   //L1: estremi (x1,y1)(x2,y2) L2: estremi (x3,y3)(x3,y3)
   int d;
   d = (x1-x2)*(y3-y4) - (y1-y2)*(x3-x4);
   if(!d)
       return 0;
   p_x = ((x1*y2-y1*x2)*(x3-x4) - (x1-x2)*(x3*y4-y3*x4))/d;
   p_y = ((x1*y2-y1*x2)*(y3-y4) - (y1-y2)*(x3*y4-y3*x4))/d;
   return 1;
}

int in_bounding_box(int x1, int y1, int x2, int y2, int p_x, int p_y)
{
    return p_x>=x1 && p_x<=x2 && p_y>=y1 && p_y<=y2;

}

int intersezione_segmenti(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4, int& p_x, int& p_y)
{
    if (!intersezione_linee(x1,y1,x2,y2,x3,y3,x4,y4,p_x,p_y))
        return 0;

    return in_bounding_box(x1,y1,x2,y2,p_x,p_y) && in_bounding_box(x3,y3,x4,y4,p_x,p_y);
}

我从《多视图几何》这本书里读到了这些算法

以下文本使用

'作为转置符号

*作为点积

当用作算子时，X作为叉乘

1. 线的定义

点x_vec = (x, y)'在直线ax + by + c = 0上

标记L = (a, b, c)'，点为(x, y, 1)'为齐次坐标

直线方程可以写成

(x, y, 1)(a, b, c)' = 0或x' * L = 0

2. 直线交点

我们有两条直线L1=(a1, b1, c1)'， L2=(a2, b2, c2)'

假设x是一个点，一个向量，x = L1 x L2 (L1叉乘L2)。

注意，x始终是一个二维点，如果你对(L1xL2)是一个三元素向量，x是一个二维坐标感到困惑，请阅读齐次坐标。

根据三重积，我们知道

L1 * (L1 x L2) = 0, L2 * (L1 x L2) = 0，因为L1,L2共平面

我们用向量x代替L1*x，那么L1*x=0, L2*x=0，这意味着x在L1和L2上，x是交点。

注意，这里x是齐次坐标，如果x的最后一个元素是零，这意味着L1和L2是平行的。

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