If the midpoints of the sides AB, BC, and CA | vedclass

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(A) Let the vertices of $	riangle ABC$ be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.Given that $D, E, F$ are midpoints of $AB, B

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$(-1, 2, -5)$

Vedclass · Answer

(A) Let the vertices of $	riangle ABC$ be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.Given that $D, E, F$ are midpoints of $AB, BC, CA$ respectively:$D = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}ight) = (1, 2, -3) \Rightarrow x_1+x_2=2, y_1+y_2=4, z_1+z_2=-6$$E = \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}, \frac{z_2+z_3}{2}ight) = (3, 0, 1) \Rightarrow x_2+x_3=6, y_2+y_3=0, z_2+z_3=2$$F = \left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}, \frac{z_1+z_3}{2}ight) = (-1, 1, -4) \Rightarrow x_1+x_3=-2, y_1+y_3=2, z_1+z_3=-8$Solving for $x_1, x_2, x_3$: $(x_1+x_2)+(x_2+x_3)+(x_1+x_3) = 2+6-2 = 6 \Rightarrow 2(x_1+x_2+x_3)=6 \Rightarrow x_1+x_2+x_3=3$. Thus $x_3=3-2=1, x_1=3-6=-3, x_2=3-(-2)=5$.Similarly for $y$: $y_1+y_2+y_3 = \frac{4+0+2}{2} = 3$. Thus $y_3=3-4=-1, y_1=3-0=3, y_2=3-2=1$.Similarly for $z$: $z_1+z_2+z_3 = \frac{-6+2-8}{2} = -6$. Thus $z_3=-6-(-6)=0, z_1=-6-2=-8, z_2=-6-(-8)=2$.So, $A(-3, 3, -8)$, $D(1, 2, -3)$, and $F(-1, 1, -4)$.The centroid of $	riangle ADF$ is $\left(\frac{-3+1-1}{3}, \frac{3+2+1}{3}, \frac{-8-3-4}{3}ight) = \left(\frac{-3}{3}, \frac{6}{3}, \frac{-15}{3}ight) = (-1, 2, -5)$.

If the midpoints of the sides $AB$, $BC$, and $CA$ of a triangle are respectively $D(1, 2, -3)$, $E(3, 0, 1)$, and $F(-1, 1, -4)$, then the centroid of the triangle $ADF$ is

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