If the points (2k, k), (k, 2k) and (k, k) wit | vedclass

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(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3

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$(8, 8)$

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(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 18$.Substituting the given points $(2k, k), (k, 2k), (k, k)$:$\frac{1}{2} |2k(2k - k) + k(k - k) + k(k - 2k)| = 18$$\frac{1}{2} |2k(k) + k(0) + k(-k)| = 18$$\frac{1}{2} |2k^2 - k^2| = 18$$\frac{1}{2} |k^2| = 18$$k^2 = 36$Since $k > 0$,we have $k = 6$.The vertices are $(12, 6), (6, 12), (6, 6)$.The centroid $(G)$ is given by $\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.$G = \left(\frac{12 + 6 + 6}{3}, \frac{6 + 12 + 6}{3}\right) = \left(\frac{24}{3}, \frac{24}{3}\right) = (8, 8)$.

If the points $(2k, k), (k, 2k)$ and $(k, k)$ with $k > 0$ enclose a triangle of area $18$ square units,then the centroid of the triangle is equal to

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