Locating the third vertex of an equilateral triangle

Question

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.

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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 $D$

Answer 1

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 $$ v= 10-4i-\varepsilon(-10+10i) = ...$$.

Answer 2

First, find the length of a side of the triangle. $$l=\sqrt{(10-0)^2 +(-4-6)^2}=10\sqrt{2}$$ Next, we draw two circles of radius $l$ and centers at the given points. The intersection of these circles will give us the third point (well, there will be two points so two solutions to this problem). We have two equations: $$(x-10)^2+(y+4)^2=200$$ and $$x^2+(y-6)^2=(x-10)^2+(y+4)^2$$ From the second equation we have $x^2+y^2-12y+36=x^2-20x+100+y^2+8y+16$ which will give us $x=y+4$. Now substitute $x$ in the first equation to get a quadratic equation for $y$. $$(y-6)^2+(y+4)^2=2y^2+4y+52=200$$ This has two solutions, $y_1=-1-5\sqrt{3}$ and $y_2=-1+5\sqrt{3}$. Thus, we have two possible locations $(3-5\sqrt{3}, -1-5\sqrt{3})$ and $(3+5\sqrt{3}, -1+5\sqrt{3})$ for the third vertex.