老实说，这就像Simon P Steven的回答一样简单，但是用这种方法，你无法控制你是否想要包含三角形边缘上的点。

我的方法有点不同，但非常基本。考虑下面的三角形;

为了在三角形中有这个点我们必须满足三个条件

ACE角(绿色)应小于ACB角(红色) ECB角(蓝色)应小于ACB角(红色) 当点E和点C的x和y值应用于|AB|直线方程时，点E和点C的符号应该相同。

在此方法中，您可以完全控制单独包含或排除边缘上的点。所以你可以检查一个点是否在三角形中，例如，只包括|AC|边。

所以我的JavaScript解决方案是这样的;

function isInTriangle(t,p){ function isInBorder(a,b,c,p){ var m = (a.y - b.y) / (a.x - b.x); // calculate the slope return Math.sign(p.y - m*p.x + m*a.x - a.y) === Math.sign(c.y - m*c.x + m*a.x - a.y); } function findAngle(a,b,c){ // calculate the C angle from 3 points. var ca = Math.hypot(c.x-a.x, c.y-a.y), // ca edge length cb = Math.hypot(c.x-b.x, c.y-b.y), // cb edge length ab = Math.hypot(a.x-b.x, a.y-b.y); // ab edge length return Math.acos((ca*ca + cb*cb - ab*ab) / (2*ca*cb)); // return the C angle } var pas = t.slice(1) .map(tp => findAngle(p,tp,t[0])), // find the angle between (p,t[0]) with (t[1],t[0]) & (t[2],t[0]) ta = findAngle(t[1],t[2],t[0]); return pas[0] < ta && pas[1] < ta && isInBorder(t[1],t[2],t[0],p); } var triangle = [{x:3, y:4},{x:10, y:8},{x:6, y:10}], point1 = {x:3, y:9}, point2 = {x:7, y:9}; console.log(isInTriangle(triangle,point1)); console.log(isInTriangle(triangle,point2));

我需要在“可控环境”中检查三角形中的点，当你绝对确定三角形是顺时针的时候。所以我拿了Perro Azul的jsfiddle，按照coproc的建议进行了修改。还去掉了多余的0.5和2乘法因为它们互相抵消了。

http://jsfiddle.net/dog_funtom/H7D7g/

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 s = (p0.y * p2.x - p0.x * p2.y + (p2.y - p0.y) * p.x + (p0.x - p2.x) * p.y); var t = (p0.x * p1.y - p0.y * p1.x + (p0.y - p1.y) * p.x + (p1.x - p0.x) * p.y); if (s <= 0 || t <= 0) return false; var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return (s + t) < A; } function checkClockwise(p0, p1, p2) { var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return A > 0; } 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() { while (true) { var result = { 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) } }; if (checkClockwise(result.a, result.b, result.c)) return result; } } <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>

以下是Unity的等效c#代码:

public static bool IsPointInClockwiseTriangle(Vector2 p, Vector2 p0, Vector2 p1, Vector2 p2)
{
    var s = (p0.y * p2.x - p0.x * p2.y + (p2.y - p0.y) * p.x + (p0.x - p2.x) * p.y);
    var t = (p0.x * p1.y - p0.y * p1.x + (p0.y - p1.y) * p.x + (p1.x - p0.x) * p.y);

    if (s <= 0 || t <= 0)
        return false;

    var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y);

    return (s + t) < A;
}

有一些恼人的边条件，即一个点恰好在两个相邻三角形的公共边上。这个点不可能在两个三角形中，也不可能不在两个三角形中。你需要一种任意但一致的方式来分配点。例如，画一条横线穿过这个点。如果这条直线与三角形的另一边在右侧相交，则该点被视为在三角形内。如果交点在左边，则该点在外面。

如果该点所在的直线是水平的，则使用above/below。

如果该点位于多个三角形的公共顶点上，则使用该点与中心点形成的角最小的三角形。

更有趣的是:三个点可以在一条直线上(零度)，例如(0,0)-(0,10)-(0,5)。在三角剖分算法中，“耳朵”(0,10)必须被切掉，生成的“三角形”是直线的退化情况。

我要做的是预先计算三个面法线，

在三维中通过边向量和面法向量的叉乘得到。 通过简单地交换分量和负一个，

对于任意一条边的内/外都是边法线和点到点向量的点积，改变符号。重复其他两(或更多)面。

好处:

在同一个三角形上进行多点测试，很多都是预先计算好的。 早期拒签的常见情况是外分多内分。(如果点分布偏向一侧，可以先测试这一侧。)

我在JavaScript中改编的高性能代码(文章如下):

function pointInTriangle (p, p0, p1, p2) {
  return (((p1.y - p0.y) * (p.x - p0.x) - (p1.x - p0.x) * (p.y - p0.y)) | ((p2.y - p1.y) * (p.x - p1.x) - (p2.x - p1.x) * (p.y - p1.y)) | ((p0.y - p2.y) * (p.x - p2.x) - (p0.x - p2.x) * (p.y - p2.y))) >= 0;
}

pointInTriangle(p, p0, p1, p2) -用于逆时针方向的三角形 pointInTriangle(p, p0, p1, p2) -用于顺时针三角形

在jsFiddle(包括性能测试)中，在一个单独的函数中也有缠绕检查。或按下面的“运行代码片段”

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(); }); document.querySelector('#performance').addEventListener('click', _testPerformance); test(); function test() { var result = checkClockwise(triangle.a, triangle.b, triangle.c) ? pointInTriangle(point, triangle.a, triangle.c, triangle.b) : pointInTriangle(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 _testPerformance () { var px = [], py = [], p0x = [], p0y = [], p1x = [], p1y = [], p2x = [], p2y = [], p = [], p0 = [], p1 = [], p2 = []; for(var i = 0; i < 1000000; i++) { p[i] = {x: Math.random() * 100, y: Math.random() * 100}; p0[i] = {x: Math.random() * 100, y: Math.random() * 100}; p1[i] = {x: Math.random() * 100, y: Math.random() * 100}; p2[i] = {x: Math.random() * 100, y: Math.random() * 100}; } console.time('optimal: pointInTriangle'); for(var i = 0; i < 1000000; i++) { pointInTriangle(p[i], p0[i], p1[i], p2[i]); } console.timeEnd('optimal: pointInTriangle'); console.time('original: ptInTriangle'); for(var i = 0; i < 1000000; i++) { ptInTriangle(p[i], p0[i], p1[i], p2[i]); } console.timeEnd('original: ptInTriangle'); } function pointInTriangle (p, p0, p1, p2) { return (((p1.y - p0.y) * (p.x - p0.x) - (p1.x - p0.x) * (p.y - p0.y)) | ((p2.y - p1.y) * (p.x - p1.x) - (p2.x - p1.x) * (p.y - p1.y)) | ((p0.y - p2.y) * (p.x - p2.x) - (p0.x - p2.x) * (p.y - p2.y))) >= 0; } function ptInTriangle(p, p0, p1, p2) { var s = (p0.y * p2.x - p0.x * p2.y + (p2.y - p0.y) * p.x + (p0.x - p2.x) * p.y); var t = (p0.x * p1.y - p0.y * p1.x + (p0.y - p1.y) * p.x + (p1.x - p0.x) * p.y); if (s <= 0 || t <= 0) return false; var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return (s + t) < A; } function render() { ctx.fillStyle = "#CCC"; ctx.fillRect(0, 0, 500, 500); drawTriangle(triangle.a, triangle.b, triangle.c); drawPoint(point); } function checkClockwise(p0, p1, p2) { var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return A > 0; } 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> <button id="performance">Run performance test (open console)</button> <pre>Click: place the point. Double click: random triangle.</pre> <pre id="result"></pre> <canvas width="500" height="500"></canvas>

受此启发: http://www.phatcode.net/articles.php?id=459

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.