Question

If coordinates of two vertices of an equilateral triangle are $(0,0)$ and $(0,2\sqrt{3}),$ then possible coordinates of the third vertex are

• A. $(3,\sqrt{3})$
• B. $(3,-\sqrt{3})$
• C. $(-3,\sqrt{3})$
• D. $(-3,-\sqrt{3})$

Answer 1

Answer: (A), (C)

$(-3,\sqrt{3})$

Let $O\equiv (0,0),A\equiv (0,2\sqrt{3})$
Let the third vertex of triangle be $B(h,k).$

Since $OAB$ is an equilateral triangle, we have
$OA=OB=AB\Rightarrow O{A}^2=O{B}^2=A{B}^2\Rightarrow 0^2+(2\sqrt{3}{)}^2=h^2+k^2=h^2+(k-2\sqrt{3}{)}^2\Rightarrow 12=h^2+k^2=h^2+k^2-4\sqrt{3}k+12\Rightarrow k=\sqrt{3},h=\pm 3$

Hence, the possible coordinates of the third vertex are $(3,\sqrt{3})$ or $(-3,\sqrt{3}).$