In the figure given below. ABC is an equilateral triangle. Find the co-ordinates of the vertices.

Explanation

From the figure, `BC =6` units.
As ABC is an equilateral triangle, `AB= BC =6` units. From A, draw `AM bot BC`, then M is the mid-point of BC.
So, `BM=(1)/(2)BC`
`(1)/(2)xx6` units=3units.
In `DeltaABM`, `angleM =90^@`. By the Pythagoras theorem, we get
`AB^2=AM^2+BM^2`
`rArr6^2-AM^2+3^2`
`rArrAM^2=36-9=27`
`rArrAM=sqrt(3)` units.
From figure, `OM=OB+BM=(1+3)` units `=4` units. Clearly, co-ordinates of B and C are `(1,0)` and `(7,0)` respectively.
As `AM=3sqrt(3)` units and `OM=4` units, therefore, co-ordinates of A are `(4,3sqrt(3))`.