If A(cos alpha, sin alpha),B(sin alpha, -cos | vedclass

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(B) Let $(h, k)$ be the centroid of the triangle. The coordinates of the centroid are given by the average of the vertices:$h = \frac{\cos \alpha + \s

Vedclass · Accepted Answer

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

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(B) Let $(h, k)$ be the centroid of the triangle. The coordinates of the centroid are given by the average of the vertices:$h = \frac{\cos \alpha + \sin \alpha + 1}{3}$ and $k = \frac{\sin \alpha - \cos \alpha + 2}{3}$Rearranging the terms,we get:$3h - 1 = \cos \alpha + \sin \alpha$ and $3k - 2 = \sin \alpha - \cos \alpha$Squaring both equations and adding them:$(3h - 1)^2 + (3k - 2)^2 = (\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2$$(9h^2 - 6h + 1) + (9k^2 - 12k + 4) = (\cos^2 \alpha + \sin^2 \alpha + 2\sin \alpha \cos \alpha) + (\sin^2 \alpha + \cos^2 \alpha - 2\sin \alpha \cos \alpha)$$9h^2 + 9k^2 - 6h - 12k + 5 = 1 + 1 = 2$$9h^2 + 9k^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 a $\Delta ABC$,then as $\alpha$ varies,the locus of its centroid is

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