Question

If two vertices of an equilateral triangle be $(0,0)$ and $(3,\sqrt{3})$, then the third vertex is ?(assuming that the triangle lies in the first quadrant)

• A. $(1,3\sqrt{3})$
• B. $(0,2\sqrt{3})$
• C. $(3,\sqrt{3})$
• D. $(1,\sqrt{3})$

Answer 1

The correct option is B $(0,2\sqrt{3})$

Let $(0,0)$ be $A$ , $(3,\sqrt{3})$ be $B$ and the third vertex be $C$

Then, angle formed by $AB$ with $x-axis$ is $ta{n}^{-1}\left(\frac{\sqrt{3}}{3}\right)={30}^{\circ }$

Since it is an equilateral triangle, $\angle CAB={60}^{\circ }$

This means that side $CA$ is perpendicular to the $x-axis$

So, vertex $C$ lies on the $y-axis$

Side length of the triangle $=\sqrt{3^2+(\sqrt{3}{)}^2}=2\sqrt{3}$

So, vertex $C$ is $(0,2\sqrt{3})$

Correct Option - (B)