Q. 24.2( 181 Votes )

# 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 :

Let ABC be the given equilateral triangle with side 2a

⇒ AB= BC = AC = 2a

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

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

**⇒ Co-ordinates of point A 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} = OA^{2} + OC^{2}

⇒(2a)^{2}= a^{2} + OC^{2}

⇒ 4a^{2} – a^{2} = OC^{2}

⇒ 3a^{2}= OC^{2}

⇒ OC =

∴ Co-ordinates of point C =

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

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

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