Question

If two vertices of an equilateral triangle an (0,0) and $(0,2\sqrt{3})$, find the third vertex.

Answer 1

Equilateral $\Delta ABCAB=BC=CAAB=\sqrt{(0-0{)}^2+(2\sqrt{3}-0{)}^2}=2\sqrt{3}AC=\sqrt{x^2+y^2}BC=x^2+(y-2\sqrt{3}{)}^{2}A{B}^2=B{C}^2=C{A}^2(2\sqrt{3}{)}^2=(\sqrt{x^2+y^2}{)}^2=[\sqrt{x^2+(y-2\sqrt{3}{)}^2}{]}^{2}12=x^2+y^2=x^2+(y-2\sqrt{3}{)}^2{x}^2+y^2=12\to \left(1\right)x^2+(y-2\sqrt{3}{)}^2=12\to (2)$

Solving equation $\left(1\right)\&\left(2\right)(x,y)=(3,\sqrt{3})$ and $(-3,\sqrt{3})$

The third vertex will be $(3,\sqrt{3})$ and $(-3,\sqrt{3})$