Creating an equilateral triangle for given two points in the plane - Python

Question

I have two points X = (x1,y1) and Y=(x2,y2) in the Cartesian plane. I need to find the third point Z = (x,y) such that these three points make an equilateral triangle.

I'm calculating the Euclidean distance between two points using the following code sample:

def distance(points, i, j):
dx = points[i][0] - points[j][0]
dy = points[i][1] - points[j][1]
return math.sqrt(dx*dx + dy*dy)

In theory, I need to equate the distances of XZ and YZ to XY.This gives us two possible answers and I need them both too. But I'm having a difficulty in initiating the point Z in my code. Can someone please help me with this? The following is a sample of what I tried.

L = [0, 6]  #known two points
d= distance(points, L[0], L[1])
x = Symbol('x')
y = Symbol('y')
newpoint = x,y   #coordintes of the third point of the triangle
f1 = distance(points, L[0], newpoint)
f2 = distance(points, L[1], newpoint)
print(nsolve((f1, f2), (x, y), (d,d)))

But this returns the following error:

 File "/Users/*.py", line 99, in <module>
    f1 = distance(points, L[0], newpoint)

  File "/Users/*.py", line 36, in distance
    dx = points[i][0] - points[j][0]

TypeError: list indices must be integers or slices, not tuple

Answer 1

In order to get the third vertex, you could just rotate the point (x2, y2) by 60 degrees around point (x1, y1). The other admissible solution would be obtained with a rotation by -60 degrees, i.e., in the opposite direction.

import math


def get_point(x1, y1, x2, y2):
    #express coordinates of the point (x2, y2) with respect to point (x1, y1)
    dx = x2 - x1
    dy = y2 - y1

    alpha = 60./180*math.pi
    #rotate the displacement vector and add the result back to the original point
    xp = x1 + math.cos( alpha)*dx + math.sin(alpha)*dy
    yp = y1 + math.sin(-alpha)*dx + math.cos(alpha)*dy

    return (xp, yp)


print(get_point(1, 1, 2, 1))
# (1.5, 0.1339745962155614)