Answer:

A` (− [tex]\sqrt{3}[/tex]a.0) , B (),a) and C ( o,-a).

Step-by-step explanation:

Let BC be the base of equilateral △ABC.

It is given that BC = 2a and mid-point of BC is at the

origin.

Therefore, coordinates of B and C are (0,a) and (0,−a) respectively.

As the triangle ABC is equilateral.

So, vertex A lies on x-axis.

Let its coordinates be (x,0).

Also, AB=BC=AC⇒AB=AC=AC=2a

Using Pythagoras theorem in ΔAOB, we get

AB ² =OA² +OB²

⇒(2a) ² =OA² +a²

⇒OA= ± [tex]\sqrt{3}[/tex]a

So, coordinates A are (− [tex]\sqrt{3}[/tex],0). Similarly, coordinates of A' are (− [tex]\sqrt{3}[/tex]a,0).

Hence ,the coordinates of the vertices of triangle are A( [tex]\sqrt{3}[/tex]a,0) , B (o, a) and C (o, -a)

or, A ` (− [tex]\sqrt{3}[/tex]a.0) , B (),a) and C ( o,-a).