39Coords 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.
