检查是一个点（x，y）是在直线上绘制的两个点之间 [英] Check is a point (x,y) is between two points drawn on a straight line

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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