我将Kris的答案移植到JavaScript。在尝试了许多不同的答案后，他给出了正确的观点。我以为我要疯了，因为我没有得到我需要的分数。

function getLineLineCollision(p0, p1, p2, p3) {
    var s1, s2;
    s1 = {x: p1.x - p0.x, y: p1.y - p0.y};
    s2 = {x: p3.x - p2.x, y: p3.y - p2.y};

    var s10_x = p1.x - p0.x;
    var s10_y = p1.y - p0.y;
    var s32_x = p3.x - p2.x;
    var s32_y = p3.y - p2.y;

    var denom = s10_x * s32_y - s32_x * s10_y;

    if(denom == 0) {
        return false;
    }

    var denom_positive = denom > 0;

    var s02_x = p0.x - p2.x;
    var s02_y = p0.y - p2.y;

    var s_numer = s10_x * s02_y - s10_y * s02_x;

    if((s_numer < 0) == denom_positive) {
        return false;
    }

    var t_numer = s32_x * s02_y - s32_y * s02_x;

    if((t_numer < 0) == denom_positive) {
        return false;
    }

    if((s_numer > denom) == denom_positive || (t_numer > denom) == denom_positive) {
        return false;
    }

    var t = t_numer / denom;

    var p = {x: p0.x + (t * s10_x), y: p0.y + (t * s10_y)};
    return p;
}

以下是对加文回答的改进。马普的解决方案也类似，但都没有推迟分割。

这实际上也是Gareth Rees的答案的一个实际应用，因为向量积在2D中的等价是补点积，这段代码用了其中的三个。切换到3D并使用叉积，在最后插入s和t，结果是3D中直线之间的两个最近点。 不管怎样，2D解:

int 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 s02_x, s02_y, s10_x, s10_y, s32_x, s32_y, s_numer, t_numer, denom, t;
    s10_x = p1_x - p0_x;
    s10_y = p1_y - p0_y;
    s32_x = p3_x - p2_x;
    s32_y = p3_y - p2_y;

    denom = s10_x * s32_y - s32_x * s10_y;
    if (denom == 0)
        return 0; // Collinear
    bool denomPositive = denom > 0;

    s02_x = p0_x - p2_x;
    s02_y = p0_y - p2_y;
    s_numer = s10_x * s02_y - s10_y * s02_x;
    if ((s_numer < 0) == denomPositive)
        return 0; // No collision

    t_numer = s32_x * s02_y - s32_y * s02_x;
    if ((t_numer < 0) == denomPositive)
        return 0; // No collision

    if (((s_numer > denom) == denomPositive) || ((t_numer > denom) == denomPositive))
        return 0; // No collision
    // Collision detected
    t = t_numer / denom;
    if (i_x != NULL)
        *i_x = p0_x + (t * s10_x);
    if (i_y != NULL)
        *i_y = p0_y + (t * s10_y);

    return 1;
}

基本上，它将除法延迟到最后一刻，并将大多数测试移动到某些计算完成之前，从而增加了早期退出。最后，它还避免了直线平行时的除零情况。

您可能还想考虑使用ε检验，而不是与零比较。非常接近平行的线会产生稍微偏离的结果。这不是一个bug，这是浮点数学的一个限制。

这是基于Gareth Ree的回答。它还返回线段重叠的情况。用c++编写的V是一个简单的向量类。其中二维中两个向量的外积返回一个标量。通过了学校自动测试系统的测试。

//Required input point must be colinear with the line
bool on_segment(const V& p, const LineSegment& l)
{
    //If a point is on the line, the sum of the vectors formed by the point to the line endpoints must be equal
    V va = p - l.pa;
    V vb = p - l.pb;
    R ma = va.magnitude();
    R mb = vb.magnitude();
    R ml = (l.pb - l.pa).magnitude();
    R s = ma + mb;
    bool r = s <= ml + epsilon;
    return r;
}

//Compute using vector math
// Returns 0 points if the lines do not intersect or overlap
// Returns 1 point if the lines intersect
//  Returns 2 points if the lines overlap, contain the points where overlapping start starts and stop
std::vector<V> intersect(const LineSegment& la, const LineSegment& lb)
{
    std::vector<V> r;

    //http://stackoverflow.com/questions/563198/how-do-you-detect-where-two-line-segments-intersect
    V oa, ob, da, db; //Origin and direction vectors
    R sa, sb; //Scalar values
    oa = la.pa;
    da = la.pb - la.pa;
    ob = lb.pa;
    db = lb.pb - lb.pa;

    if (da.cross(db) == 0 && (ob - oa).cross(da) == 0) //If colinear
    {
        if (on_segment(lb.pa, la) && on_segment(lb.pb, la))
        {
            r.push_back(lb.pa);
            r.push_back(lb.pb);
            dprintf("colinear, overlapping\n");
            return r;
        }

        if (on_segment(la.pa, lb) && on_segment(la.pb, lb))
        {
            r.push_back(la.pa);
            r.push_back(la.pb);
            dprintf("colinear, overlapping\n");
            return r;
        }

        if (on_segment(la.pa, lb))
            r.push_back(la.pa);

        if (on_segment(la.pb, lb))
            r.push_back(la.pb);

        if (on_segment(lb.pa, la))
            r.push_back(lb.pa);

        if (on_segment(lb.pb, la))
            r.push_back(lb.pb);

        if (r.size() == 0)
            dprintf("colinear, non-overlapping\n");
        else
            dprintf("colinear, overlapping\n");

        return r;
    }

    if (da.cross(db) == 0 && (ob - oa).cross(da) != 0)
    {
        dprintf("parallel non-intersecting\n");
        return r;
    }

    //Math trick db cross db == 0, which is a single scalar in 2D.
    //Crossing both sides with vector db gives:
    sa = (ob - oa).cross(db) / da.cross(db);

    //Crossing both sides with vector da gives
    sb = (oa - ob).cross(da) / db.cross(da);

    if (0 <= sa && sa <= 1 && 0 <= sb && sb <= 1)
    {
        dprintf("intersecting\n");
        r.push_back(oa + da * sa);
        return r;
    }

    dprintf("non-intersecting, non-parallel, non-colinear, non-overlapping\n");
    return r;
}

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。

我将Kris的答案移植到JavaScript。在尝试了许多不同的答案后，他给出了正确的观点。我以为我要疯了，因为我没有得到我需要的分数。

function getLineLineCollision(p0, p1, p2, p3) {
    var s1, s2;
    s1 = {x: p1.x - p0.x, y: p1.y - p0.y};
    s2 = {x: p3.x - p2.x, y: p3.y - p2.y};

    var s10_x = p1.x - p0.x;
    var s10_y = p1.y - p0.y;
    var s32_x = p3.x - p2.x;
    var s32_y = p3.y - p2.y;

    var denom = s10_x * s32_y - s32_x * s10_y;

    if(denom == 0) {
        return false;
    }

    var denom_positive = denom > 0;

    var s02_x = p0.x - p2.x;
    var s02_y = p0.y - p2.y;

    var s_numer = s10_x * s02_y - s10_y * s02_x;

    if((s_numer < 0) == denom_positive) {
        return false;
    }

    var t_numer = s32_x * s02_y - s32_y * s02_x;

    if((t_numer < 0) == denom_positive) {
        return false;
    }

    if((s_numer > denom) == denom_positive || (t_numer > denom) == denom_positive) {
        return false;
    }

    var t = t_numer / denom;

    var p = {x: p0.x + (t * s10_x), y: p0.y + (t * s10_y)};
    return p;
}

C和Objective-C

基于Gareth Rees的回答

const AGKLine AGKLineZero = (AGKLine){(CGPoint){0.0, 0.0}, (CGPoint){0.0, 0.0}};

AGKLine AGKLineMake(CGPoint start, CGPoint end)
{
    return (AGKLine){start, end};
}

double AGKLineLength(AGKLine l)
{
    return CGPointLengthBetween_AGK(l.start, l.end);
}

BOOL AGKLineIntersection(AGKLine l1, AGKLine l2, CGPoint *out_pointOfIntersection)
{
    // http://stackoverflow.com/a/565282/202451

    CGPoint p = l1.start;
    CGPoint q = l2.start;
    CGPoint r = CGPointSubtract_AGK(l1.end, l1.start);
    CGPoint s = CGPointSubtract_AGK(l2.end, l2.start);
    
    double s_r_crossProduct = CGPointCrossProductZComponent_AGK(r, s);
    double t = CGPointCrossProductZComponent_AGK(CGPointSubtract_AGK(q, p), s) / s_r_crossProduct;
    double u = CGPointCrossProductZComponent_AGK(CGPointSubtract_AGK(q, p), r) / s_r_crossProduct;
    
    if(t < 0 || t > 1.0 || u < 0 || u > 1.0)
    {
        if(out_pointOfIntersection != NULL)
        {
            *out_pointOfIntersection = CGPointZero;
        }
        return NO;
    }
    else
    {
        if(out_pointOfIntersection != NULL)
        {
            CGPoint i = CGPointAdd_AGK(p, CGPointMultiply_AGK(r, t));
            *out_pointOfIntersection = i;
        }
        return YES;
    }
}

CGFloat CGPointCrossProductZComponent_AGK(CGPoint v1, CGPoint v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

CGPoint CGPointSubtract_AGK(CGPoint p1, CGPoint p2)
{
    return (CGPoint){p1.x - p2.x, p1.y - p2.y};
}

CGPoint CGPointAdd_AGK(CGPoint p1, CGPoint p2)
{
    return (CGPoint){p1.x + p2.x, p1.y + p2.y};
}

CGFloat CGPointCrossProductZComponent_AGK(CGPoint v1, CGPoint v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}

CGPoint CGPointMultiply_AGK(CGPoint p1, CGFloat factor)
{
    return (CGPoint){p1.x * factor, p1.y * factor};
}

许多函数和结构都是私有的，但是你应该很容易就能知道发生了什么。 这是公开的在这个回购https://github.com/hfossli/AGGeometryKit/