For the variable triangle ABC with fixed vertex (1,2) and A,B having coordinates (cost,sint),(sint,-cost) respectively, Find the locus of its centroid.

x = (a +b +c)/3 , y = (d + e + f)/3

So coordinate of given triangle is :

```
h = (cost +sint +1)/3 , k = (sint - cost + 2 )/3
```

So 3h - 1 = cost + sint (1)

And 3k -2 = sint - cost (2)

So squaring and adding both we have

(3h - 1)^{2} + (3k -2)^{2} = (cost + sint)^{2} + (sint - cost)^{2}

9h^{2} +1 -6h + 9k^{2} + 4 -12k = 2

9h^{2} +3 -6h + 9k^{2} -12k = 0

Replacing h and k by x and y , we have the locus eqaution

9x^{2} +3 -6x + 9y^{2} -12y = 0

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