The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point

Let us consider ABC be the given equilateral triangle with side 2a.

Where, AB = BC = AC = 2a

In the above figure, by assuming that the base BC lies on the x axis such that the mid-point of BC is at the origin i.e. BO = OC = a, where O is the origin.

The co-ordinates of point C are (0, a) and that of B are (0,-a)

Since the line joining a vertex of an equilateral ∆ with the mid-point of its opposite side is perpendicular. So, vertex A lies on the y –axis

By applying Pythagoras theorem

(AC)² = OA² + OC²

(2a)²= a² + OC²

4a² – a² = OC²

3a² = OC²

OC =\( \sqrt{3a}\)

Co-ordinates of point C = ± \( \sqrt{3a}\), 0

The vertices of the given equilateral triangle are (0, a), (0, -a), (\( \sqrt{3a}\), 0)

Or (0, a), (0, -a) and (\( - \sqrt{3a}\), 0)