If the origin is the centroid of the triangle | vedclass

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(D) The centroid 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 $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+

Vedclass · Accepted Answer

$p=1, q=3, r=-2$

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(D) The centroid 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 $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.Given that the origin $(0, 0, 0)$ is the centroid:For the $x$-coordinate: $\frac{2+q-5}{3} = 0$ $\Rightarrow q-3 = 0$ $\Rightarrow q = 3$.For the $y$-coordinate: $\frac{p-2+1}{3} = 0$ $\Rightarrow p-1 = 0$ $\Rightarrow p = 1$.For the $z$-coordinate: $\frac{-3+5+r}{3} = 0$ $\Rightarrow r+2 = 0$ $\Rightarrow r = -2$.Thus,$p=1, q=3, r=-2$.

If the origin is the centroid of the triangle whose vertices are $A(2, p, -3)$,$B(q, -2, 5)$,and $C(-5, 1, r)$,then

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