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(A) Let the centroid of $	riangle ABC$ be $(x, y)$.The coordinates of the vertices are $A(\cos \alpha, \sin \alpha)$,$B(\sin \alpha, -\cos \alpha)$,a

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the centroid of $	riangle ABC$ be $(x, y)$.The coordinates of the vertices are $A(\cos \alpha, \sin \alpha)$,$B(\sin \alpha, -\cos \alpha)$,and $C(1, 2)$.The centroid $(x, y)$ is given by:$x = \frac{\cos \alpha + \sin \alpha + 1}{3} \implies 3x - 1 = \cos \alpha + \sin \alpha$$y = \frac{\sin \alpha - \cos \alpha + 2}{3} \implies 3y - 2 = \sin \alpha - \cos \alpha$Squaring both equations:$(3x - 1)^2 = (\cos \alpha + \sin \alpha)^2 = \cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha = 1 + 2 \sin \alpha \cos \alpha$$(3y - 2)^2 = (\sin \alpha - \cos \alpha)^2 = \sin^2 \alpha + \cos^2 \alpha - 2 \sin \alpha \cos \alpha = 1 - 2 \sin \alpha \cos \alpha$Adding the two squared equations:$(3x - 1)^2 + (3y - 2)^2 = 1 + 2 \sin \alpha \cos \alpha + 1 - 2 \sin \alpha \cos \alpha = 2$$9x^2 - 6x + 1 + 9y^2 - 12y + 4 = 2$$9x^2 + 9y^2 - 6x - 12y + 3 = 0$Dividing by $3$:$3(x^2 + y^2) - 2x - 4y + 1 = 0$.

If $A(\cos \alpha, \sin \alpha)$,$B(\sin \alpha, -\cos \alpha)$,and $C(1, 2)$ are the vertices of a $\triangle ABC$,then the locus of its centroid is:

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