Find the third point of a regular triangle?

Question

Geometry

Knowing the coordinates of point 1(x1,y1) and the coordinates of point 2(x2,y2) find the third point(x3,y3) of a regular triangle with side d. Accordingly, the angles of an equilateral triangle are equal.
There are such thoughts:
x3=|(x2-x1)/2| + (d*sqrt(3)) / 2* sin(a)
the problem is just to find the angle (a), there were guesses that it is equal to = 90 degrees - arcsine (|y2-y1|/d)

Answer 1

Your logic is correct - take the middle of the segment AB and set aside a perpendicular from it with a long sqrt (3) / 2 * d.
But you don't need to look for angles, the perpendicular vector is trivially found - this is {y2-y1, x1-x2} (You can prove perpendicularity through the scalar product, for example). Moreover, the length of this vector will be already d (after all, this is a vector rotated by 90 degrees along the side of the triangle). Means it remains to be stupidly multiplied by sqrt(3)/2.
Thus the formula x3 = (x1+x2)/2 +sqrt(3)/2*(y2-y1).

Answer 2

Knowing the coordinates of point 1(x1,y1) and the coordinates of point 2(x2,y2) find the third point(x3,y3) of a regular triangle with side d.

Illiterate wording. Not points, but vertices. d is redundant.
If A(x1,y1), B(x2,y2), then the third vertex C(x3,y3) is found by rotating vertex B around A by 60 degrees clockwise and counterclockwise.