Locus of centroid of a $\triangle{ABC}$ with $A=(\cos\alpha,\sin\alpha)$, $B=(\sin\alpha,-\cos\alpha)$, and $C=(1,2)$

Question

If $A(\cos\alpha, \sin\alpha)$, $B(\sin\alpha, -\cos\alpha)$ and $C(1,2)$ are the vertices of $\triangle ABC$, find the locus of the triangle's centroid as $\alpha$ varies.

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Let centroid be $(h,k)$, $$(h,k)\equiv\left(\frac{\sin(\alpha) + \cos(\alpha) + 1}{3},\frac{\sin(\alpha) - \cos(\alpha) + 2}{3}\right)\tag{1}$$ How to proceed further? Hints would be appreciated!

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As suggested by @wonderman

From $(1)$, $$\sin(\alpha) + \cos(\alpha) = 3h - 1 \qquad \sin(\alpha) - \cos(\alpha) = 3k - 2$$ When we square the above and add, $$ (3h - 1)^2 + (3k - 1)^2 = 2$$ Substituting $h$ as $x$ and $k$ as $y$, $$3x^2 + 3y^2 - 2h - 4k + 1 = 0$$

Answer 1

Consider $(\sin \alpha + \cos \alpha)^2 + (\sin \alpha - \cos \alpha)^2$. What does this work out to?

Answer 2

The centroid of $ABC$ lies on the segment joining $C$ with the midpoint $M$ of $AB$, precisely at $\frac{2}{3}$ of such segment. In your problem $M$ travels from $(-1,0)$ to $(1,0)$ and $C$ is fixed, so the locus of centroids is a segment with endpoints $(-1/3,2/3)$ and $(1,2/3)$.