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(N/A) The centroid $(G)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula: $G = \left( \frac{x_1 + x_2 + x_3}{

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(N/A) The centroid $(G)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula: $G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.
Given the vertices $(1, a), (2, b), (c^2, -3)$,the $x$-coordinate of the centroid is $x_G = \frac{1 + 2 + c^2}{3} = \frac{3 + c^2}{3} = 1 + \frac{c^2}{3}$.
For the centroid to lie on the $Y$-axis,its $x$-coordinate must be $0$.
Setting $x_G = 0$,we get $1 + \frac{c^2}{3} = 0$,which implies $\frac{c^2}{3} = -1$,or $c^2 = -3$.
Since the square of any real number $c$ cannot be negative $(c^2 \geq 0)$,there is no real value of $c$ for which $x_G = 0$.
Therefore,the centroid cannot lie on the $Y$-axis.

Prove that the centroid of the triangle with vertices $(1, a), (2, b), (c^2, -3)$ does not lie on the $Y$-axis.

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