CO-ORDINATE

Coords of vertices of triangle are A(3, 1), B(0, 1 + √3) and C(0, 1 - √3).
Lengths are a = AB = √9 + 3 = √12
b = BC = √12
c = AC = √12
Thus this is an equilateral triangle.

The circumcentre coincides with the centroid luckily here. Otherwise it is a point which is equidistant from all sides of the triangle.

So our circumcentre is (1,1).

I'm not good in rotation of axes...but I'll try. Since we are rotating the axes wrt circumcentre, it is our origin. Our shift of origin is from (1,1) to (0,0). (To make rotation easier)
0 = X, 0 = Y.
1 = x, 1 = y.
X = x+h, Y = y+k
Thus our new circumcentre is (-1, -1).
Similarly the vertices become (2, 0) , (-1, √3) and (-1, -√3).

Angle of rotation = +60°(in ACW direction).

Let us take point A to rotate.
The old point is (x,y) = (2,0) [As per our new origin]
x = Xcos60 - Ysin60
y = Xsin60 + Ycos60
where (X, Y) are the new coords.
2 = X/2 - Y√3/2
0 = X√3/2 + Y/2
Multiply by 2,
4 = X - √3Y
0 = √3X + Y

or

4√3 = X√3 - 3Y
0 = 3√3X + 3Y

Adding,
4√3 = 4√3X or X = 1, Y = -√3
So A(1, -√3). Add 1,1 to shift back to old origin to get (2, 1 - √3).
Similarly do the rest? lola. There might be a shorter method but...this is all that occurred to me so sorry.

the new vertices are (-1,1) (2,1-√3) (2,1+√3)is this correct

array use rotation theorem

A(3+i), B i(1 + √3) and C i(1 - √3).

1+i is ur circumcenter and points A B C lie on a circle with center z₀=1+i

new rotated positions wrt to center will also lie on same circle

now u hav to rotate A B C wrt to center

let Z_Abe new rotated vertex of A

so from coni theorem

Z_A-(1+i)(3+i)-(1+i )=e^iπ/3

similarly do for oders.

coni theorem is der in the arihant algebra pritish

@pritish

its not something extraordinary...

u can see it in any coaching material.......btw der is a whole exercise on this theorem in MLk..

aur agar isko nai pada toh samajhlo complex nahi pada. ;)

i cud find this link on net see it - http://bit.ly/ae9HpJ