Question

The base of an equilateral triangle with side 2a lies along they 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.
Accordingly, AB = BC = CA = 2a
Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).
It is known 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 the y-axis.

On applying Pythagoras theorem to ΔAOC, we obtain
(AC)² = (OA)² + (OC)²
⇒ (2a)² = (OA)² + a²
⇒ 4a² – a² = (OA)²
⇒ (OA)² = 3a²
⇒ OA = √ 3a

∴Coordinates of point A =  (+√3a,0 
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and ( √ 3a,0 )
or (0, a), (0, –a), and   ( - √ 3a,  0 )