python 已知平行四边形三个点,求第四个点的案例

import numpy as np
#已知平行四边形三个点，求第四个点
#计算两点之间的距离
def CalcEuclideanDistance(point1,point2):
  vec1 = np.array(point1)
  vec2 = np.array(point2)
  distance = np.linalg.norm(vec1 - vec2)
  return distance
#计算第四个点
def CalcFourthPoint(point1,point2,point3): #pint3为A点
  D = (point1[0]+point2[0]-point3[0],point1[1]+point2[1]-point3[1])
  return D
#三点构成一个三角形，利用两点之间的距离，判断邻边AB和AC,利用向量法以及平行四边形法则，可以求得第四个点D
def JudgeBeveling(point1,point2,point3):
  dist1 = CalcEuclideanDistance(point1,point2)
  dist2 = CalcEuclideanDistance(point1,point3)
  dist3 = CalcEuclideanDistance(point2,point3)
  dist = [dist1, dist2, dist3]
  max_dist = dist.index(max(dist))
  if max_dist == 0:
    D = CalcFourthPoint(point1,point2,point3)
  elif max_dist == 1:
    D = CalcFourthPoint(point1,point3,point2)
  else:
    D = CalcFourthPoint(point2,point3,point1)
  return D
 
print(JudgeBeveling((0,1),(1,0),(1,1)))
print(JudgeBeveling((5,39),(500,35),(496,17)))

def line_slope(x1,y1,x2,y2,x3,y3,x4,y4):
    k1=(y2-y1)/(x2-x1)
    k2=(y3-y2)/(x3-x2)
    k3=(y4-y3)/(x4-x3)
    k4=(y1-y4)/(x1-x4)
    return k1,k2,k3,k4

        l1=int(tk1*(i-tx1)+ty1)
        l2=int(tk2*(i-tx2)+ty2)
        l3=int(tk3*(i-tx3)+ty3)
        l4=int(tk4*(i-tx4)+ty4)