Question

Find the centroid of a triangle, mid-points of whose sides are (1, 2, -3), (3, 0, 1) and (-1, 1, -4)

Answer 1

A $(x_1,y_1,z_1),B(x_2,y_2,z_2)$ and C $(x_3,y_3,z_3)$be the

vertices of the given triangle and let

D(1, 2, -3), E(3, 0, 1) and F(-1, 1, -4) be the mid-point of the sides

BC, CA and AB respectively.

D is the mid point respectively.

D is the mid-point of BC.

$\therefore \frac{x_2+x_3}{2}=1,\frac{y_2+y_3}{2}=2,\frac{z_2+z_3}{2}=-3$

$\Rightarrow x_2+x_3=2,y_2+y_3=4,z_2+z_3$ = - 6 (i)

E is the mid-point of CA.

$\therefore \frac{x_2+x_3}{2}=3,\frac{y_1+y_3}{2}=0,\frac{z_1+z_3}{2}=1$

$\Rightarrow x_1+x_3=3,y_1+y_3=0,z_1+z_3$ = 2 (ii)

F is the mid-point of AB.

$\therefore \frac{x_1+x_2}{2}=-1,\frac{y_1+y_3}{2}=1,\frac{z_1+z_2}{2}=-4$

$\Rightarrow x_1+x_2=-2,y_1+y_2=2,z_1+z_2$=-8(iii)

Adding the first three equations we get,

$2(x_1+x_2+x_3)=6,2(y_1+y_2+y_3)$ = 6 and 2$(z_1+z_2+z_3)=-12$

$\Rightarrow x_1+x_2+x_3=3,y_1+y_2+y_3$ = 3 and

$z_1+z_2+z_3=-6$

The coordinate of the centroid dis given by

$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3})$

$\Rightarrow$ (1, 1, -2)