Question

If two vertices of an equilateral triangle are $(3,0)$ and $(6,0)$, find the third vertex.

Answer 1

$A(3,0)$ and $B(6,0)$, let third vertex $C(a,b)$

$=>A{B}^2={BC}^2={CA}^2=>{\left(3\right)}^2+{\left(0\right)}^2={(3-a)}^2+b^2={(6-a)}^2+b^2=>a^2-6a+9+b^2=36+a^2-12a+b^2=9=>a=\frac{27}{6}=>b={\left(\frac{3}{2}\right)}^2-(3{)}^2=>b=-\frac{27}{4}$

So, $C(\frac{9}{2},-\frac{27}{4})$