查找海拔与基准相交的点(Python) [英] Find point where altitude meets base (Python)

问题描述

仅给出一个锐角三角形顶点的坐标，我如何有效而快速地找到某个特定顶点的高度与相反的底点相交的点的坐标?

Given only the coordinates of the vertices of an acute triangle, how can I efficiently and quickly find the coordinates of the point at which the altitude from a particular vertex meets the opposite base?

仅使用数学，numpy或scipy的解决方案将非常有用.

A solution using only math, numpy, or scipy would be incredibly helpful.

推荐答案

需要的点是顶点(例如顶点C)在包含相反面(例如AB)的线上的正交投影.

Needed point is orthogonal projection of vertex point (say vertex C) onto the line containing opposite side (say AB).

要找到投影点，请获取AB和AC的向量

To find projection point, get vectors for AB and AC

 AB = (B - A)    //in coordinates ab.x = b.x-a.x, ab.y = b.y-a.y
 AC = (C - A)

并使用AB和AC的标量积查找参数

and find parameter using scalar product of AB and AC

t =(AB * AC) / (AB * AB) 
t =((b.x-a.x)*(c.x-a.x) + (b.y-a.y)*(c.y-a.y)) / ((b.x-a.x)*(b.x-a.x) + (b.y-a.y)*(b.y-a.y))

投影点坐标

 P = A + AB * t
 p.x = a.x + (b.x-a.x) * t
 p.y = a.y + (b.y-a.y) * t

仅此而已

def orthoProjection(ax, ay, bx, by, cx, cy):
    abx = bx - ax
    aby = by - ay
    acx = cx - ax
    acy = cy - ay
    t = (abx * acx + aby * acy) / (abx * abx + aby * aby)
    px = ax + t * abx
    py = ay + t * aby
    return px, py

print(orthoProjection(0, 0, 4, 4, -1, 5))
>>(2.0, 2.0)

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