Question

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

Answer 1

O(0,0) and A$(3,\sqrt{3})$ be the given points and let B(x,y) be the third vertex of equilateral $△OAB$. Then,

$OA=OB=AB$

$\Rightarrow {OA}^2={OB}^2={AB}^2$

$Wehave,{OA}^2={(3-0)}^2+{(\sqrt{3}-0)}^2=12,$
${OB}^2=x^2+y^2$
$and,{AB}^2={(x-3)}^2+{(y-\sqrt{3})}^2$
$\Rightarrow {AB}^2=x^2+y^2-6x-2\sqrt{3}y+12$
$\therefore {OA}^2={OB}^2={AB}^2$
$\Rightarrow {OA}^2={OB}^{2}and{OB}^2={AB}^2$
$\Rightarrow x^2+y^2=12$
$and,x^2+y^2=x^2+y^2-6x-2\sqrt{3}y+12$
$\Rightarrow x^2+y^2=12and6x+2\sqrt{3}y=12$
$\Rightarrow x^2+y^2=12and3x+\sqrt{3}y=6$
$\Rightarrow x^2+{\left(\frac{6-3x}{\sqrt{3}}\right)}^2=12[\because 3x+\sqrt{3}y=6\therefore y=\frac{6-3x}{\sqrt{3}}]$
$\Rightarrow {3x}^2+{(6-3x)}^2=12$
$\Rightarrow {12x}^2-36x=0$
$\Rightarrow x=0,3$
$\therefore x=0\Rightarrow \sqrt{3}y=6\Rightarrow y=\frac{6}{\sqrt{3}}=2\sqrt{3}[Puttingx=0in3x+\sqrt{3}y=6]$
$and,x=3\Rightarrow 9+\sqrt{3}y=6\Rightarrow y=\frac{6-9}{\sqrt{3}}=-\sqrt{3}[Puttingx=3in3x+\sqrt{3}y=6]$
Hence, the coordinates of the third vertex B are $(0,2\sqrt{3})or,(3,-\sqrt{3})$.