if `A(cosalpha,sinalpha),B(sinalpha,-cosalpha),C(1,2)` are the vertices of Triangle ABC, Find the locsus of its centroid.

Explanation

Let (h,k) be the triangle . Then, `h=(cosalpha+sinalpha+1)/(3)`
and `k=(sinalpha-cosalpha+2)/(3)`
or `3h-1=cosalpha+sinalpha`
and `3k-2=sinalpha-cosalpha`
Squareing and adding, we get
` (3h-1)^2+(3k-2)^2=2`
or `9(h^2+k^2)-6h-12k+3=0`
or `3(h^2+k^2)-2h-4k+1=0`
Therefore the locus of the centroid is `3(x^2+y^2)-2x-4y+1=0` .