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

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

Distance between (0,0) and$(3,\sqrt3)$ is

$d=\sqrt{3^2+\sqrt3^2}=\sqrt{12}=2\sqrt3$

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, $(x-3)^2+(y-\sqrt3)^2=12$

$\Rightarrow x^2-6x+9+y^2-2\sqrt3y+3=12$

$\Rightarrow (x^2+y^2)-6x-2\sqrt3y=0$

$\Rightarrow (12)-6x-2\sqrt3y=0$ (from (i))

$\Rightarrow 3x+\sqrt3y=6$

$\Rightarrow x=\cfrac{6-\sqrt3y}{3}$

on putting this value in (i), we get

$(\cfrac{6-\sqrt3y}{3})^2+y^2=12$ ...(ii)

$\Rightarrow (\cfrac{36+3y^2-12\sqrt3y}{9})+y^2=12$

$\Rightarrow 12y^2-12\sqrt3y=72$

$\Rightarrow y^2-\sqrt3y-6=0$

$\Rightarrow y^2-2\sqrt3y+\sqrt3y-6=0$

$\Rightarrow y(y-2\sqrt3)+\sqrt3(y-2\sqrt3)=0$

$\Rightarrow y=2\sqrt3,-\sqrt3$

form (ii)

when, $y=-\sqrt3;x=3$

$y=2\sqrt3;x=0$

Therefore, points are $(0,2\sqrt3)$ or $(3,-\sqrt3)$