Question

The base of an equilateral triangle with side $2a$ lies along the $y$-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Answer 1

Let $ABC$ be the given equilateral triangle with side $2a$

$\Rightarrow AB=BC=CA=2a$

Since base, $BC$ lies on the $y$-axis with mid-point at origin $O$.

$\Rightarrow OB=OC=a$

So, the coordinates of point $B$ are $(0,a)$ and the coordinates of $C$ are $(0,-a)$

We know that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex $A$ lies on $x$-axis.

So, let the coordinates of point $A$ be $(x,0)$

On applying Pythagoras theorem $△ABC$, we get

$(AC{)}^2=(OA{)}^2+(OC{)}^2$

$\Rightarrow (2a{)}^2=x^2+a^2$

$\Rightarrow 4a^2-a^2=x^2$

$\Rightarrow x^2=3a^2$

$\Rightarrow x=\pm \sqrt{3a}$

$\therefore$ coordinates of point $A$ are $(\pm \sqrt{3a},0)$ or $(-\sqrt{3a},0)$

Thus, the vertices of the given equilateral triangle are $(0,a)(0,-a)$ and $(\sqrt{3a},0)$ or $(0,a)(0,-a)$ and $(-\sqrt{3a},0)$