If the centroid of Delta ABC is G(3, 3) and t | vedclass

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(A) Let the coordinates of vertex $A$ be $(x, y)$.The centroid $G(x_g, y_g)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given

Vedclass · Accepted Answer

$\frac{69}{2}$

Vedclass · Answer

(A) Let the coordinates of vertex $A$ be $(x, y)$.The centroid $G(x_g, y_g)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$.Given $G(3, 3)$,$B(-1, 2)$,and $C(2, -3)$,we have:$3 = \frac{x - 1 + 2}{3} \implies 9 = x + 1 \implies x = 8$.$3 = \frac{y + 2 - 3}{3} \implies 9 = y - 1 \implies y = 10$.So,vertex $A$ is $(8, 10)$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |8(2 - (-3)) + (-1)(-3 - 10) + 2(10 - 2)|$.Area $= \frac{1}{2} |8(5) + (-1)(-13) + 2(8)|$.Area $= \frac{1}{2} |40 + 13 + 16| = \frac{1}{2} |69| = \frac{69}{2}$ square units.

If the centroid of $\Delta ABC$ is $G(3, 3)$ and the vertices are $B(-1, 2)$ and $C(2, -3)$,then find the area of $\Delta ABC$.

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