Question

对于一条直线，斜率k是已知的，一个点（x1，y1）是已知的，如何使用python得到另一个点（x2，y2）？我知道计算的方法，但不知道使用python进行编码。

（Y2-Y1）/（X2-X1）= K
sqrt（（x2-x1）^ 2 +（y2-y1）^ 2）= length

我试着简化如下，但似乎我不能应用linalg-solve

kx2-y2-kx1 + y1 = 0
（y2-y1）^ 2 +（x2-x1）^ 2 = length ^ 2

补充： 感谢你的所有答案... Posh_Pumpkin的代码正是我想要的，以前我认为我需要应用我经常使用的linalg-solve。

这是基于他的答案的测试代码：假设P1 =（1,1）p2 =（x，y），p1p2 = sqrt（2），k = 1，那么p2必须是=（2,2）

import numpy as np
import math
k = 1
d = math.sqrt(2)
p1 = (1,1)
r_sq = d**2 / (1 + k**2)
r = math.sqrt(r_sq)
p2 = (p1[0] + r, p1[1] + k*r)
print(p2)

Answer 1

为什么要使用复杂的解决方案？我会在P1 = (0, 0)。

的假设下接近它

假设P1位于原点，并且知道slope = k，您知道P2 = (r, k*r)为常数r。可以通过计算距离r来计算P1P2。既然你说你已经知道了距离，我们可以简单地做：

r^2 + (k*r)^2 = d^2

查找r。找到r后，您可以在P2时获得P1 = (0, 0)的坐标。因此，要查找实际坐标，您只需执行P2 = P1 + P2。检查以下内容：

import math

# Obviously assume k and d are known constants, and P1 is a known point
k = # given k
d = # given d
p1 = # (given x coord, given y coord)

r_sq = d**2 / (1 + k**2)
r = math.sqrt(r_sq)
p2 = (p1[0] + r, p1[1] + k*r)
print(p2)

如果你必须使用numpy，你应该告诉我们你的代码以及究竟令你烦恼的是什么。什么阻止您使用该模块，无论是错误还是意外行为？

Answer 2

您可以使用基本触发来解决此问题。这是一般推导。

let p1 = (x1,y1) & p2 = (x2 = x1+d, y2 = y1+h), 
let L be the distance between p1 & p2 
* note for p1 & p2 such that x1 != x2 && y1 != y2, a triangle can be formed Ldh such that tan(theta) = h/d

h/d is the slope of the line (m) connecting points p1 & p2, so tan(theta) = m
 => theta = atan(m), from the law of sines ( sin(a)/A = sin(b)/B ) 
 => sin(90)/L = sin(atan(theta))/y2
 => y2 = L*sin( atan(theta) )

now get x from the point slope form of a line y= y1+m(x-x1) = (y-y1)/m +x1
so x2 = (y2-y1)/m + x1

这是用python表达的：

from math import sin,  atan
from random import randint

# This is the formula for (x2,y2) = p2
x = lambda y2, m, x1, y1: (y2 - y1)/float(m) + x1
y = lambda l, m, y1: l*sin( atan(m)  ) + y1

# p2 constraints ( x2 > x1, y2 > y1 or x2 > x1, y2 < y1 )
p1 =  [randint(1,1000),randint(1,1000)]
p2 =  [randint(p1[0],1001),randint(0,p1[1])]

# calculate distance between p1 & p2 (L), also calculate slope (m)
slope = lambda x1,y1,x2,y2: (y2-y1)/float(x2-x1)
dist  = lambda x1,y1,x2,y2: ( (y2-y1)**2 + (x2-x1)**2  )**(0.5)
L = dist(p1[0],p1[1],p2[0],p2[1])
m = slope(p1[0],p1[1],p2[0],p2[1])

#  now see if our functions for x & y yield p2
y2 = y(L,m,p1[1])
p_derived = [ x(y2,m,p1[0],p1[1]),y2  ]
print "p1: ",p1 , "p2 actual: ",p2, "p2 derived: ",p_derived

所以在这里我生成两个随机点p1和p2，并通过将我的派生结果与实际结果进行比较来验证p2是否可以从斜率，距离和p1计算出来。

Python：直线，Slope k已知，一点P1已知，长度P1P2已知，如何获得P2？

2 个答案: