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.

Question

Philoid · Accepted Answer

Let ABC be the given equilateral triangle with side 2a

⇒ AB= BC = AC = 2a

Assuming that the base BC lies on the y axis such that the mid-point of BC is at the origin

i.e.BO =OC = a and O is the origin

⇒ Co-ordinates of point 0 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

⇒ Vertex of A lies on the y –axis

On applying Pythagoras theorem

(AC)² = OA² + OC²

⇒(2a)²= OA² + a²

⇒ 4a² – a² = OA²

⇒ 3a²= OA²

⇒ OA =

∴ Co-ordinates of point A =

Thus, the vertices of the given equilateral triangle are (0, a) , (0, -a) , (

Or (0, a), (0, -a) and (