Given two points calculate a third point in a way that three point construct an equilateral triangle.

Question

So I have two points lets say A(x1,y1) and B(x2,y2). I want to find a point C (there will be two points) in which if you connect the points you will have an equilateral triangle. I know that if I draw a circle from each point with radius of equal to AB I will find the point C at the intersection of these two circles. But I don't know how to calculate that point.

Answer 1

Comment turned to answer per request.

Instead of drawing circles and find the intersection, one can rotate $B$ with respect to $A$ for $\pm 60^\circ$ to get $C(x_3,y_3)$. In matrix notation, the formula is:

$$\begin{bmatrix}x_3\\ y_3\end{bmatrix} = \begin{bmatrix}x_1\\ y_1\end{bmatrix} + \begin{bmatrix}\frac12 & \mp \frac{\sqrt{3}}{2}\\ \pm\frac{\sqrt{3}}{2} & \frac12\end{bmatrix} \begin{bmatrix}x_2-x_1\\ y_2-y_1\end{bmatrix} $$

Answer 2

Let $C(x_o, y_o)$ $$AB=\sqrt{((x_1-x_2)^2+(y_1-y_2)^2}$$ The following must happen: $$AC=BC=AB$$ Or else: $$AC=AB \ and \ BC=AB$$ And you solve the two equations.

Answer 3

Let $r=AB=\sqrt{((x_1-x_2)^2+(y_1-y_2)^2}$. The equation of the circle radius of $AB$ and $A$ as the center will be $(x-x_1)^2+(y-y_1)^2=r^2$. The equation of the circle radius of $AB$ and $B$ as the center will be $(x-x_2)^2+(y-y_2)^2=r^2$.

Thus you need to solve $(x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2$. Simplifying this we'll get $-2x_1x+x_1^2-2y_1y+y_1^2=-2x_2x+x_1^2-2y_2y+y_2^2$.
Now substitute $y=y_1+\sqrt{r^2-(x-x_1)^2}$ and solve the resulting equation to find $x$