Определить координаты 4 вершины квадрата

import math
 
x1,y1 = map(int,input().split())
x2,y2 = map(int,input().split())
x3,y3 = map(int,input().split())
sumx1 = x1-x2
sumx = []
sumy = []
sumx2 = x1-x3
sumx3 = x2-x3
sumy1 = math.sqrt(sumx1**2 + (y1-y2)**2)
sumy2 = math.sqrt(sumx3**2 + (y1-y3)**2)
sumy3 = math.sqrt(sumx2**2 + (y2-y3)**2)
sumx.append(sumy1)
sumx.append(sumy2)
sumx.append(sumy3)
k = sumx.index(max(sumx))
if k ==0:
    h = x1-x3+x2
    k = y1-y3+y2
elif k == 1:
    h = x1-x2+x3
    k = y1-y2+y3
elif k == 2:
    h = x2- x1+x3
    k = y2-y1+y3
print(h,k)

x1,y1 = map(int,input().split())
x2,y2 = map(int,input().split())
x3,y3 = map(int,input().split())
x4 = set((y1,y2,y3)) - set((x1,x2,x3))
y4 = set((x1,x2,x3)) - set((y1,y2,y3))
print(*x4, *y4)

x1,y1 = map(int,input().split())
x2,y2 = map(int,input().split())
x3,y3 = map(int,input().split())
 
# для скалярного произведения векторов
v12 = (x1-x2) * (y1-y2)
v13 = (x1-x3) * (y1-y3)
v23 = (x2-x3) * (y2-y3)
 
if v12 == -v13:
    x4 = x2 + x3 - x1
    y4 = y2 + y3 - y1
elif v12 == -v23:
    x4 = x1 + x3 - x2
    y4 = y1 + y3 - y2
else:
    x4 = x1 + x2 - x3
    y4 = y1 + y2 - y3
    
print((x4, y4))

import numpy as np
from scipy.spatial.distance import pdist
points=np.array([[float(input(f'{i}{j}=')) for i in 'xy']for  j in '₁₂₃'])
#points=np.array([[2.0, 1.0], [5.0, -1.0], [4.0, 4.0]])
edges_dist=pdist(points)[::-1]
# the last point opposite the long side of the triangle:
ordered_points=points[np.argsort(edges_dist), :] 
# homogeneous coordinates:
ordered_points_homogeneous=np.hstack((ordered_points, np.ones((3,1))))
# translate matrix
T=np.array([[1,0,0], [0,1,0],[-ordered_points[0,0],-ordered_points[0,1],1]])
# rotate angle
angle=np.arctan2((ordered_points_homogeneous@T)[1,1], (ordered_points_homogeneous@T)[1,0])
# rotate matrix:
R=np.array([[np.cos(angle),-np.sin(angle),0], [np.sin(angle),np.cos(angle),0],[0,0,1]])
# mirror matrix
M=np.array([[1,0,0], [0,-1,0],[0,0,1]])
# all transform matrix:
A=T@R@M@np.linalg.inv(R)@np.linalg.inv(T)
# fourth vertex of the square:
pos=(ordered_points_homogeneous[2,:]@A)[:-1]
print(f'x₄={pos[0]:.2f}, y₄={pos[1]:.2f}')

points=[[float(input(f'{i}{j}=')) for i in 'xy']for  j in '₁₂₃']
#points=[[2.0, 1.0], [5.0, -1.0], [4.0, 4.0]]
mass_center=[sum(i)/3 for i in zip(*points)] # calculate mass center
points2=[[points[j][i]-mass_center[i] for i in (0,1)] for j in (0,1,2)] # origin to mass center
length=[(j, x[0]*x[0]+x[1]*x[1]) for j, x in enumerate(points2)] # system of vector from mass center to vertices
idx=min(length, key=lambda x: x[1])[0] # index shortest vector from mass center
points2.append(points2.pop(idx)) # shortest vector to end list
fourth_vertex=[x[0]+x[1]-x[2] for x in zip(*points2)]
pos=[x[0]+x[1] for x in zip(mass_center, fourth_vertex)] #reconstruct origin
print(f'x₄={pos[0]:.2f}, y₄={pos[1]:.2f}')