你如何在距离一条线的给定垂直距离上找到一个点？ [英] How do you find a point at a given perpendicular distance from a line?

问题描述

我在窗口中画了一条线，然后让用户拖动它。所以，我的线由两点定义：（x1，y1）和（x2，y2）。但是现在我想在我的线条的末端绘制帽子，也就是说，在我的每个端点上都有短垂线。因此，为了在终点（x1，y1）绘制我的帽子线，我需要找到两个点，它们组成一个垂直线，并且其每个点都离开点（x1，y1）N / 2个像素点。所以如何计算一个点（x3，y3） ），因为它需要与已知线的终点（x1，y1）相距一个垂直距离N / 2，即由（x1，y1）和（x2，y2）定义的直线？

解决方案

您需要计算垂直于线段的单位矢量。避免计算斜率，因为这可能导致除以零错误。

  dx = x1-x2 
 dy = y1 -y2 
 dist = sqrt（dx * dx + dy * dy）
 dx / = dist 
 dy / = dist 
 x3 = x1 +（N / 2）* dy 
 y3 = y1  - （N / 2）* dx 
 x4 = x1  - （N / 2）* dy 
 y4 = y1 +（N / 2）* dx 
  




I have a line that I draw in a window and I let the user drag it around. So, my line is defined by two points: (x1,y1) and (x2,y2). But now I would like to draw "caps" at the end of my line, that is, short perpendicular lines at each of my end points. The caps should be N pixels in length.

Thus, to draw my "cap" line at end point (x1,y1), I need to find two points that form a perpendicular line and where each of its points are N/2 pixels away from the point (x1,y1).

So how do you calculate a point (x3,y3) given it needs to be at a perpendicular distance N/2 away from the end point (x1,y1) of a known line, i.e. the line defined by (x1,y1) and (x2,y2)?

解决方案

You need to compute a unit vector that's perpendicular to the line segment. Avoid computing the slope because that can lead to divide by zero errors.

dx = x1-x2
dy = y1-y2
dist = sqrt(dx*dx + dy*dy)
dx /= dist
dy /= dist
x3 = x1 + (N/2)*dy
y3 = y1 - (N/2)*dx
x4 = x1 - (N/2)*dy
y4 = y1 + (N/2)*dx

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