<s>[INST] Locating the third vertex of an equilateral triangle Two vertices of an equilateral triangle are at $A=(10,-4)$ and $B=(0,6)$. 
How can one locate the third vertex?
Maybe someone could give me the easy way please.

My attempt: 


*

*Find the average of $M=(x , y)$ of $A$ and $B$, which is $(10+0)/2 = 5$, $(-4+6)/2 = 1$. So M(5,1)$. 

*Find the equation of the perpendicular bisector. Slope of $AB = 10/-10 = -1$, Slope $m$ of the perpendicular bisector $= +1$
$$ \Rightarrow y-1 = 1*(x-5)\Rightarrow  y = x-4 .$$

*There are $2$ vertices, $C$ and $D$, both on the line $y = x-4$. The altitude of the triangle is $10$. Find the $2$ points on $y = x-4$ which is $10$ units distance from $M(5,1)$:
$$d^2 = (\text{difference of } x)^2 + (\text{difference of } y)^2 \Rightarrow 100 = (x-5)^2 + (y-1)^2$$ 
Sub for $y = x-4$:
$$ 100 = (x-5)^2 + (x-5)^2\Rightarrow (x-5)^2 = 50 \Rightarrow x-5 = \pm\sqrt{50}$$ 
Thus $x = 5 + \sqrt{50}$, $y = 1 + \sqrt{50}$ and so the Vertices are $C$: $x = 5 - \sqrt{50}$, $y = 1 - \sqrt{50}$ --> Vertex $D$
 [/INST] We know that multiplying with $\varepsilon ={1\over 2}+{\sqrt{3}\over 2}$ is rotation around $0$ for $60^{\circ}$ in counterclockwise direction and with $-\varepsilon$ around $0$ for $60^{\circ}$ in clockwise direction.
Let in complex plain $z=10-4i$ correspond to the point $(10,-4)$ and $w=6i$ to the point $(0,6)$. If $v$ is third point of this triangle then $v-z = \varepsilon(w-z)$ (if we rotate $w$ around $z$ for $60^{\circ}$ in counterclockwise direction) or $v-z = -\varepsilon(w-z)$ (if we rotate $w$ around $z$ for $60^{\circ}$ in clockwise direction). In first case we have 
$$v= 10-4i+\varepsilon(-10+10i) = ...$$ and in second case we have $$ v= 10-4i-\varepsilon(-10+10i) = ...$$.  
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