Question

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

Answer 1

Two vertices of an equilateral triangle are $(0,0)and(3,\surd 3).$

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

Distance between $(0,0)and(x,y)$ = Distance between$(0,0)and(3,\surd 3)$ = Distance between $(x,y)and(3,\surd 3)$

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

$x^2+y^2=12$

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

$24-6x-2\sqrt{3y}=12$

$-6x-2\sqrt{3y}=-12$

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

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

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

$\frac{(36+3y2-12\sqrt{3y})}{9}+y^2=12$

$36+3y^2-12\sqrt{3y}+9y2=108$

$-12\sqrt{3y}+12y^2-72=0$

$\sqrt{-3y}+y2-6=0$

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

$y=2\sqrt{3}or\sqrt{-3}$

$Ify=2\sqrt{3},x=(6-6)/3=0$

$Ify=-\sqrt{3},x=\frac{9}{3}=3$

So,the third vertex of the equilateral triangle$=(0,2\sqrt{3})or(3,-\sqrt{3}).$