The points (11,9), (2,1) and (2,-1) are the m | vedclass

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(D) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Given mid-points are $F(11, 9)$,$E(2, 1)$,and $D(2, -1)$.Since t

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$(5,3)$

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(D) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Given mid-points are $F(11, 9)$,$E(2, 1)$,and $D(2, -1)$.Since the centroid of a triangle formed by joining the mid-points of the sides of a triangle is the same as the centroid of the original triangle,we can directly calculate it.The centroid $(G)$ of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.Let the mid-points be $(x_a, y_a) = (11, 9)$,$(x_b, y_b) = (2, 1)$,and $(x_c, y_c) = (2, -1)$.The centroid of the triangle formed by these mid-points is $\left(\frac{11+2+2}{3}, \frac{9+1-1}{3}\right) = \left(\frac{15}{3}, \frac{9}{3}\right) = (5, 3)$.Thus,the centroid of the original triangle is $(5, 3)$.

The points $(11,9), (2,1)$ and $(2,-1)$ are the mid-points of the sides of a triangle. Then,the centroid is

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