If the origin is the centroid of the triangle | vedclass

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(C) The centroid $(x, y, z)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by: $x = \frac{x_1+x_2+x_3

Vedclass · Accepted Answer

$-2, -\frac{16}{3}, 2$

Vedclass · Answer

(C) The centroid $(x, y, z)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by: $x = \frac{x_1+x_2+x_3}{3}$,$y = \frac{y_1+y_2+y_3}{3}$,$z = \frac{z_1+z_2+z_3}{3}$ Since the origin $(0, 0, 0)$ is the centroid,we have: $0 = \frac{2a - 4 + 8}{3} \Rightarrow 2a + 4 = 0 \Rightarrow a = -2$ $0 = \frac{2 + 3b + 14}{3} \Rightarrow 3b + 16 = 0 \Rightarrow b = -\frac{16}{3}$ $0 = \frac{6 - 10 + 2c}{3} \Rightarrow 2c - 4 = 0 \Rightarrow c = 2$ Thus,the values are $a = -2, b = -\frac{16}{3}, c = 2$.

If the origin is the centroid of the triangle $PQR$ with vertices $P(2a, 2, 6)$,$Q(-4, 3b, -10)$,and $R(8, 14, 2c)$,then the values of $a, b, c$ respectively are:

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