If G(3, -5, r) is the centroid of triangle AB | vedclass

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(A) The centroid $G$ 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{

Vedclass · Accepted Answer

-$2$,$3$,$2$

Vedclass · Answer

(A) The centroid $G$ 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 $A(7, -8, 1)$,$B(p, q, 5)$,$C(q+1, 5p, 0)$,and $G(3, -5, r)$.Equating the coordinates:$3 = \frac{7 + p + q + 1}{3} \implies 9 = 8 + p + q \implies p + q = 1 \quad (1)$$-5 = \frac{-8 + q + 5p}{3} \implies -15 = -8 + q + 5p \implies 5p + q = -7 \quad (2)$$r = \frac{1 + 5 + 0}{3} = \frac{6}{3} = 2$Subtracting $(1)$ from $(2)$:$(5p + q) - (p + q) = -7 - 1$$4p = -8 \implies p = -2$Substituting $p = -2$ in $(1)$:$-2 + q = 1 \implies q = 3$Thus,$p = -2, q = 3, r = 2$.

If $G(3, -5, r)$ is the centroid of $\triangle ABC$,where $A \equiv (7, -8, 1)$,$B \equiv (p, q, 5)$,and $C \equiv (q+1, 5p, 0)$ are vertices of the triangle $ABC$,then the values of $p, q, r$ are respectively:

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