使用JavaScript在Triangle中查找第三点 [英] Finding 3rd point in Triangle using JavaScript

问题描述

好的，所以我需要在JavaScript中弄清楚如何在三角形上绘制第三个点。参见下图。

Ok so I need to figure out in JavaScript how to plot a third point on a triangle. See diagram below.

A和B将是随机点（根据它们相对于0,0的原点而定，可能为正或负）。

A and B will be randomized points (may be positive or negative depending on where they fall relative to the 0,0 origin.)

因此，A和B是已知点（x和y）。

thus A and B are known points (x and y).

我已经弄清楚如何根据A绘制C

I already figured out how to plot C based on A and B.

C和DI之间的距离要控制。例如，我想说 C和D之间的距离现在是20px ... D在哪里？

The distance between C and D I want control over. for example I want to say "The distance between C and D is now 20px... where is D?"

所以我想举个例子，我们可以说C& D之间的距离将为20px。那意味着我有CD和CB，但没有DB。我也知道C（x，y）和B（x，y）。

So I guess for example purposes we can say that the distance between C&D will be 20px. That means I have CD and CB, but not DB. I also know C(x,y) and B(x,y).

我现在需要找到D ...我不是数学家，所以请像我5岁时向我解释一下。 ，尝试使用多个示例，但我仍然迷路...例如：我看到提到了一些带有phi的方程..什么是phi？

I need to find D now... I'm not a math mind so please explain it to me like I'm 5. I've googled it multiple times, attempted to use multiple examples, and I'm still lost... For example: I saw some equations with phi mentioned.. what is phi? How do I use phi in JavaScript terms, etc...

摘要：

A(x,y) is known (randomized)
B(x,y) is known (randomized)
C(x,y) is known (midpoint of AB)
CB is known (using distance formula)
CD = 20 pixels (or whatever I set it to).
DB = ???
D(x,y) = ???

这是我到目前为止的内容，但这可能是错误的。。

Here's what I have so far, but it's probably wrong..

var Aeh = {x:50, y:75};
    var Bee = {x:300, y:175};
    var Cee = {x:0, y:0};
    var Dee = {x:0, y:0};

    window.onload = function(){ 
         refreshPoints();
         solveForC();
    } 
    function refreshPoints(){
        TweenLite.set("#pointA", {x:Aeh.x, y:Aeh.y});
        TweenLite.set("#pointB", {x:Bee.x, y:Bee.y});
        TweenLite.set("#pointC", {x:Cee.x, y:Cee.y});
        TweenLite.set("#pointD", {x:Dee.x, y:Dee.y});
    }
    function solveForC() {
        Cee.x = (Bee.x + Aeh.x)/2;
        Cee.y = (Bee.y + Aeh.y)/2;
        refreshPoints();
        solveForD();
    }
    function solveForD() {
        // Dee.x = AB * Cos(Φ) + x_1
        // Dee.y = AB * Sin(Φ) + y_1

        Dee.x = (Cee.x+Bee.x/2) * Math.cos((1 + Math.sqrt(5)) / 2) + Cee.x;
        Dee.y = (Cee.y+Bee.y/2) * Math.sin((1 + Math.sqrt(5)) / 2) + Cee.y;

        refreshPoints();
    }

推荐答案

您有A，B和C（中点），您知道从C到D的距离已设置（例如20），并且从C到D与从A到B的直线成直角。使用此方法可以找到D的两个解决方案。理解的最简单方法是找到从A到B的线的角度，并使用该角度来计算D。

You have A, B and C (the midpoint), you know the distance from C to D is set (say at 20), and the line from C to D is at right angles to the line from A to B. You can find two solutions for D using this. The simplest way to understand is to find the angle of the line from A to B, and use that help calculate D.

var angleAB = Math.atan2(B.y - A.y, B.x - A.x);
// to get the angle of the line from C to D, add 90 degrees
// in radians, that is Math.PI / 2
var angleCD = angleAB + Math.PI / 2;

// now you can calculate one of D's solutions
// the 20 represents your distance from C to D, and can be changed if desired.
DeeOne.x = C.x + 20 * Math.cos(angleCD);
DeeOne.y = C.y + 20 * Math.sin(angleCD);

// a second solution can be found by going in the other direction from C
DeeTwo.x = C.x - 20 * Math.cos(angleCD);
DeeTwo.y = C.x - 20 * Math.sin(angleCD);

如果您需要的只是一定距离（例如， ）。希望这会有所帮助。

There might be ways to cut out some of this calculation if all you need is a certain distance (etc) from the diagram. Hope this helps.

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