FWIW，下面的函数(在C中)既检测线的交点，又确定交点。这是基于Andre LeMothe的“Tricks of the Windows Game Programming Gurus”中的一个算法。这与其他答案(例如Gareth的答案)中的一些算法并没有什么不同。然后LeMothe使用克莱默法则(不要问我)来解这些方程。

我可以证明它在我的小行星克隆中起作用，并且似乎正确地处理了Elemental, Dan和Wodzu在其他答案中描述的边缘情况。它也可能比KingNestor发布的代码快，因为它都是乘法和除法，没有平方根!

我想这里有一些除以0的可能性，尽管在我的例子中这不是问题。很容易修改以避免崩溃。

// Returns 1 if the lines intersect, otherwise 0. In addition, if the lines 
// intersect the intersection point may be stored in the floats i_x and i_y.
char get_line_intersection(float p0_x, float p0_y, float p1_x, float p1_y, 
    float p2_x, float p2_y, float p3_x, float p3_y, float *i_x, float *i_y)
{
    float s1_x, s1_y, s2_x, s2_y;
    s1_x = p1_x - p0_x;     s1_y = p1_y - p0_y;
    s2_x = p3_x - p2_x;     s2_y = p3_y - p2_y;

    float s, t;
    s = (-s1_y * (p0_x - p2_x) + s1_x * (p0_y - p2_y)) / (-s2_x * s1_y + s1_x * s2_y);
    t = ( s2_x * (p0_y - p2_y) - s2_y * (p0_x - p2_x)) / (-s2_x * s1_y + s1_x * s2_y);

    if (s >= 0 && s <= 1 && t >= 0 && t <= 1)
    {
        // Collision detected
        if (i_x != NULL)
            *i_x = p0_x + (t * s1_x);
        if (i_y != NULL)
            *i_y = p0_y + (t * s1_y);
        return 1;
    }

    return 0; // No collision
}

顺便说一句，我必须说，在LeMothe的书中，虽然他显然得到了正确的算法，但他展示的具体示例插入了错误的数字，并且计算错误。例如:

(4 * (4-1) + 12 * (7-1))/(17 * 4 + 12 * 10) = 844/0.88 = 0.44

这让我困惑了好几个小时。：(

一个c++程序，用于检查两条给定线段是否相交

#include <iostream>
using namespace std;

struct Point
{
    int x;
    int y;
};

// Given three colinear points p, q, r, the function checks if
// point q lies on line segment 'pr'
bool onSegment(Point p, Point q, Point r)
{
    if (q.x <= max(p.x, r.x) && q.x >= min(p.x, r.x) &&
        q.y <= max(p.y, r.y) && q.y >= min(p.y, r.y))
       return true;

    return false;
}

// To find orientation of ordered triplet (p, q, r).
// The function returns following values
// 0 --> p, q and r are colinear
// 1 --> Clockwise
// 2 --> Counterclockwise
int orientation(Point p, Point q, Point r)
{
    // See 10th slides from following link for derivation of the formula
    // http://www.dcs.gla.ac.uk/~pat/52233/slides/Geometry1x1.pdf
    int val = (q.y - p.y) * (r.x - q.x) -
              (q.x - p.x) * (r.y - q.y);

    if (val == 0) return 0;  // colinear

    return (val > 0)? 1: 2; // clock or counterclock wise
}

// The main function that returns true if line segment 'p1q1'
// and 'p2q2' intersect.
bool doIntersect(Point p1, Point q1, Point p2, Point q2)
{
    // Find the four orientations needed for general and
    // special cases
    int o1 = orientation(p1, q1, p2);
    int o2 = orientation(p1, q1, q2);
    int o3 = orientation(p2, q2, p1);
    int o4 = orientation(p2, q2, q1);

    // General case
    if (o1 != o2 && o3 != o4)
        return true;

    // Special Cases
    // p1, q1 and p2 are colinear and p2 lies on segment p1q1
    if (o1 == 0 && onSegment(p1, p2, q1)) return true;

    // p1, q1 and p2 are colinear and q2 lies on segment p1q1
    if (o2 == 0 && onSegment(p1, q2, q1)) return true;

    // p2, q2 and p1 are colinear and p1 lies on segment p2q2
    if (o3 == 0 && onSegment(p2, p1, q2)) return true;

     // p2, q2 and q1 are colinear and q1 lies on segment p2q2
    if (o4 == 0 && onSegment(p2, q1, q2)) return true;

    return false; // Doesn't fall in any of the above cases
}

// Driver program to test above functions
int main()
{
    struct Point p1 = {1, 1}, q1 = {10, 1};
    struct Point p2 = {1, 2}, q2 = {10, 2};

    doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

    p1 = {10, 0}, q1 = {0, 10};
    p2 = {0, 0}, q2 = {10, 10};
    doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

    p1 = {-5, -5}, q1 = {0, 0};
    p2 = {1, 1}, q2 = {10, 10};
    doIntersect(p1, q1, p2, q2)? cout << "Yes\n": cout << "No\n";

    return 0;
}

iMalc回答的Python版本:

def find_intersection( p0, p1, p2, p3 ) :

    s10_x = p1[0] - p0[0]
    s10_y = p1[1] - p0[1]
    s32_x = p3[0] - p2[0]
    s32_y = p3[1] - p2[1]

    denom = s10_x * s32_y - s32_x * s10_y

    if denom == 0 : return None # collinear

    denom_is_positive = denom > 0

    s02_x = p0[0] - p2[0]
    s02_y = p0[1] - p2[1]

    s_numer = s10_x * s02_y - s10_y * s02_x

    if (s_numer < 0) == denom_is_positive : return None # no collision

    t_numer = s32_x * s02_y - s32_y * s02_x

    if (t_numer < 0) == denom_is_positive : return None # no collision

    if (s_numer > denom) == denom_is_positive or (t_numer > denom) == denom_is_positive : return None # no collision


    # collision detected

    t = t_numer / denom

    intersection_point = [ p0[0] + (t * s10_x), p0[1] + (t * s10_y) ]


    return intersection_point

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;
}

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

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。