Question

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

• A. $(0,2\sqrt{3})$
• B. $(0,-\sqrt{3})$
• C. $(3,-\sqrt{3})$
• D. $(3,2\sqrt{3})$

Answer 1

Answer: (A), (C)

The correct options are
A $(0,2\sqrt{3})$
C $(3,-\sqrt{3})$

Let (x,y) be the third vertex of the triangle.

Distance between (0,0) and$(3,\sqrt{3})$ is

$d=\sqrt{3^2+{\sqrt{3}}^2}=\sqrt{12}=2\sqrt{3}$

Since the triangle is equilateral, therefore all the sides are equal.

$\Rightarrow (x-0{)}^2+(y-0{)}^2=12$

$\Rightarrow x^2+y^2=12$ ...(i)

Again, $(x-3{)}^2+(y-\sqrt{3}{)}^2=12$

$\Rightarrow x^2-6x+9+y^2-2\sqrt{3}y+3=12$

$\Rightarrow (x^2+y^2)-6x-2\sqrt{3}y=0$

$\Rightarrow \left(12\right)-6x-2\sqrt{3}y=0$ (from (i))

$\Rightarrow 3x+\sqrt{3}y=6$

$\Rightarrow x=\frac{6-\sqrt{3}y}{3}$

on putting this value in (i), we get

$(\frac{6-\sqrt{3}y}{3}{)}^2+y^2=12$ ...(ii)

$\Rightarrow \left(\frac{36+3y^2-12\sqrt{3}y}{9}\right)+y^2=12$

$\Rightarrow 12y^2-12\sqrt{3}y=72$

$\Rightarrow y^2-\sqrt{3}y-6=0$

$\Rightarrow y^2-2\sqrt{3}y+\sqrt{3}y-6=0$

$\Rightarrow y(y-2\sqrt{3})+\sqrt{3}(y-2\sqrt{3})=0$

$\Rightarrow y=2\sqrt{3},-\sqrt{3}$

form (ii)

when, $y=-\sqrt{3};x=3$

$y=2\sqrt{3};x=0$

Therefore, points are $(0,2\sqrt{3})$ or $(3,-\sqrt{3})$