问题描述
我画了两点之间的一条线A（x，y）--- B（x，y）
现在我有第三个点C（x，y）。我想知道如果C位于A和B之间的线上。
我想用java语言编写。我找到了几个类似的答案。但是，所有都有一些问题，没有人是完美的。
I have drawn a line between two points A(x,y)---B(x,y)Now I have a third point C(x,y). I want to know that if C lies on the line which is drawn between A and B.I want to do it in java language. I have found couple of answers similar to this. But, all have some problems and no one is perfect.
推荐答案
if (distance(A, C) + distance(B, C) == distance(A, B))
return true; // C is on the line.
return false; // C is not on the line.
或只是：
return distance(A, C) + distance(B, C) == distance(A, B);
这种方式非常简单。如果C位于 AB 行，您将得到以下情形：
The way this works is rather simple. If C lies on the AB line, you'll get the following scenario:
A-C------B
并且，无论它在哪一行， dist（AC）+ dist（CB）== dist（AB）。对于任何其他情况，你有一个描述的三角形和'dist（AC）+ dist（CB）> dist（AB）'：
and, regardless of where it lies on that line, dist(AC) + dist(CB) == dist(AB). For any other case, you have a triangle of some description and 'dist(AC) + dist(CB) > dist(AB)':
A-----B
\ /
\ /
C
事实上，如果C位于外推线上，这甚至有效：
In fact, this even works if C lies on the extrapolated line:
C---A-------B
只要距离保持无符号。距离 dist（AB）的计算公式如下：
provided that the distances are kept unsigned. The distance dist(AB) can be calculated as:
___________________________
/ 2 2
V (A.x - B.x) + (A.y - B.y)
请记住浮点运算的固有限制（有限精度）。您可能需要选择足够接近的测试（例如，小于百万分之一的错误）以确保相等的正确运行。
Keep in mind the inherent limitations (limited precision) of floating point operations. It's possible that you may need to opt for a "close enough" test (say, less than one part per million error) to ensure correct functioning of the equality.
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