In the given figure, ABC is an equilateral triangle. Find the co-ordinates of A.

Given:

The co-ordinates of B = (2, 0)

The co-ordinates of C = (6, 0)

So, the length of BC = 6 - 2 = 4 units

Since ABC is an equilateral triangle,

The height of the triangle = $\dfrac{\sqrt3}{2}a$

= $\dfrac{\sqrt3}{2} \times 4$

= 2 $\sqrt3$ units

The mid-point of BC = $\Big(\dfrac{2+6}{2}, \dfrac{0+0}{2}\Big)$

= (4, 0)

Since Δ ABC is equilateral, and BC lies on the x-axis:

• The abscissa (x-coordinate) of A is the same as the midpoint of BC, i.e., x=4.

• The ordinate (y-coordinate) of A is the height of the triangle, $\sqrt{3}$.

Hence, the co-ordinates of A are (4, 2 $\sqrt{3}$).