一个简单的方法是:

找出连接 分别指向三角形的三个点 顶点和夹角之和 这些向量。如果它们的和 角度是2*那么点是 在三角形里面。

两个解释替代方案的好网站是:

黑卒和沃尔夫勒姆

最简单的方法，适用于所有类型的三角形，就是确定P点A点B点C点的角。如果任何一个角大于180.0度，那么它在外面，如果是180.0度，那么它在圆周上，如果acos欺骗了你，小于180.0度，那么它在里面。看一看理解http://math-physics-psychology.blogspot.hu/2015/01/earlish-determination-that-point-is.html

通过使用重心坐标的解析解(由Andreas Brinck指出)和:

不是把乘法分布在括号里的项上 通过存储相同的项来避免多次计算 还原比较(如coproc和Thomas Eding所指出的)

可以最小化“昂贵”操作的数量:

function ptInTriangle(p, p0, p1, p2) {
    var dX = p.x-p2.x;
    var dY = p.y-p2.y;
    var dX21 = p2.x-p1.x;
    var dY12 = p1.y-p2.y;
    var D = dY12*(p0.x-p2.x) + dX21*(p0.y-p2.y);
    var s = dY12*dX + dX21*dY;
    var t = (p2.y-p0.y)*dX + (p0.x-p2.x)*dY;
    if (D<0) return s<=0 && t<=0 && s+t>=D;
    return s>=0 && t>=0 && s+t<=D;
}

代码可以粘贴在Perro Azul jsfiddle中，或者通过点击下面的“运行代码片段”来尝试

var ctx = $("canvas")[0].getContext("2d"); var W = 500; var H = 500; var point = { x: W / 2, y: H / 2 }; var triangle = randomTriangle(); $("canvas").click(function(evt) { point.x = evt.pageX - $(this).offset().left; point.y = evt.pageY - $(this).offset().top; test(); }); $("canvas").dblclick(function(evt) { triangle = randomTriangle(); test(); }); test(); function test() { var result = ptInTriangle(point, triangle.a, triangle.b, triangle.c); var info = "point = (" + point.x + "," + point.y + ")\n"; info += "triangle.a = (" + triangle.a.x + "," + triangle.a.y + ")\n"; info += "triangle.b = (" + triangle.b.x + "," + triangle.b.y + ")\n"; info += "triangle.c = (" + triangle.c.x + "," + triangle.c.y + ")\n"; info += "result = " + (result ? "true" : "false"); $("#result").text(info); render(); } function ptInTriangle(p, p0, p1, p2) { var A = 1/2 * (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); var sign = A < 0 ? -1 : 1; var s = (p0.y * p2.x - p0.x * p2.y + (p2.y - p0.y) * p.x + (p0.x - p2.x) * p.y) * sign; var t = (p0.x * p1.y - p0.y * p1.x + (p0.y - p1.y) * p.x + (p1.x - p0.x) * p.y) * sign; return s > 0 && t > 0 && (s + t) < 2 * A * sign; } function render() { ctx.fillStyle = "#CCC"; ctx.fillRect(0, 0, 500, 500); drawTriangle(triangle.a, triangle.b, triangle.c); drawPoint(point); } function drawTriangle(p0, p1, p2) { ctx.fillStyle = "#999"; ctx.beginPath(); ctx.moveTo(p0.x, p0.y); ctx.lineTo(p1.x, p1.y); ctx.lineTo(p2.x, p2.y); ctx.closePath(); ctx.fill(); ctx.fillStyle = "#000"; ctx.font = "12px monospace"; ctx.fillText("1", p0.x, p0.y); ctx.fillText("2", p1.x, p1.y); ctx.fillText("3", p2.x, p2.y); } function drawPoint(p) { ctx.fillStyle = "#F00"; ctx.beginPath(); ctx.arc(p.x, p.y, 5, 0, 2 * Math.PI); ctx.fill(); } function rand(min, max) { return Math.floor(Math.random() * (max - min + 1)) + min; } function randomTriangle() { return { a: { x: rand(0, W), y: rand(0, H) }, b: { x: rand(0, W), y: rand(0, H) }, c: { x: rand(0, W), y: rand(0, H) } }; } <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> <pre>Click: place the point. Double click: random triangle.</pre> <pre id="result"></pre> <canvas width="500" height="500"></canvas>

导致:

变量“召回”:30 可变存储:7 补充:4 减法:8 乘法:6 部门:没有 比较:4

这与Kornel Kisielewicz解决方案(25次召回，1次存储，15次减法，6次乘法，5次比较)相比非常好，如果需要顺时针/逆时针检测(它本身需要6次召回，1次加法，2次减法，2次乘法和1次比较，使用解析解行列式，如rhgb所指出的)，可能会更好。

一般来说，最简单(也是最优)的算法是检查由边创建的半平面的哪一边是点。

以下是关于GameDev的一些高质量信息，包括性能问题。

这里有一些代码让你开始:

float sign (fPoint p1, fPoint p2, fPoint p3)
{
    return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}

bool PointInTriangle (fPoint pt, fPoint v1, fPoint v2, fPoint v3)
{
    float d1, d2, d3;
    bool has_neg, has_pos;

    d1 = sign(pt, v1, v2);
    d2 = sign(pt, v2, v3);
    d3 = sign(pt, v3, v1);

    has_neg = (d1 < 0) || (d2 < 0) || (d3 < 0);
    has_pos = (d1 > 0) || (d2 > 0) || (d3 > 0);

    return !(has_neg && has_pos);
}

If you know the co-ordinates of the three vertices and the co-ordinates of the specific point, then you can get the area of the complete triangle. Afterwards, calculate the area of the three triangle segments (one point being the point given and the other two being any two vertices of the triangle). Thus, you will get the area of the three triangle segments. If the sum of these areas are equal to the total area (that you got previously), then, the point should be inside the triangle. Otherwise, the point is not inside the triangle. This should work. If there are any issues, let me know. Thank you.

bool point2Dtriangle(double e,double f, double a,double b,double c, double g,double h,double i, double v, double w){
    /* inputs: e=point.x, f=point.y
               a=triangle.Ax, b=triangle.Bx, c=triangle.Cx 
               g=triangle.Ay, h=triangle.By, i=triangle.Cy */
    v = 1 - (f * (b - c) + h * (c - e) + i * (e - b)) / (g * (b - c) + h * (c - a) + i * (a - b));
    w = (f * (a - b) + g * (b - e) + h * (e - a)) / (g * (b - c) + h * (c - a) + i * (a - b));
    if (*v > -0.0 && *v < 1.0000001 && *w > -0.0 && *w < *v) return true;//is inside
    else return false;//is outside
    return 0;
} 

从质心转换而来的几乎完美的笛卡尔坐标 在*v (x)和*w (y)双精度内导出。 在每种情况下，两个导出双精度对象前面都应该有一个*字符，可能是*v和*w 代码也可以用于四边形的另一个三角形。 特此签名只写三角形abc从顺时针abcd的四边形。

A---B
|..\\.o|  
|....\\.| 
D---C 

o点在ABC三角形内 对于带有第二个三角形的测试，将此函数称为CDA方向，*v=1-*v后的结果应正确;* w = 1 - * w;为了四合院