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Question

(C) Let the centroid of the triangle be $(h, k)$.Then,$h = \frac{\cos \alpha + \sin \alpha + 1}{3}$ and $k = \frac{\sin \alpha - \cos \alpha + 2}{3}$.

Vedclass · Accepted Answer

$3(x^2 + y^2) - 2x - 4y + 1 = 0$

Vedclass · Answer

(C) Let the centroid of the triangle be $(h, k)$.Then,$h = \frac{\cos \alpha + \sin \alpha + 1}{3}$ and $k = \frac{\sin \alpha - \cos \alpha + 2}{3}$.This implies $3h - 1 = \cos \alpha + \sin \alpha$ and $3k - 2 = \sin \alpha - \cos \alpha$.Squaring and adding both equations:$(3h - 1)^2 + (3k - 2)^2 = (\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2$.$(3h - 1)^2 + (3k - 2)^2 = (\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha) + (\sin^2 \alpha + \cos^2 \alpha - 2 \sin \alpha \cos \alpha) = 1 + 1 = 2$.$9h^2 - 6h + 1 + 9k^2 - 12k + 4 = 2$.$9(h^2 + k^2) - 6h - 12k + 3 = 0$.Dividing by $3$,we get $3(h^2 + k^2) - 2h - 4k + 1 = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $3(x^2 + y^2) - 2x - 4y + 1 = 0$.

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

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