Question

Find the third vertex of triangle whose centroid is origin and two vertices are $(2,4,6)$ and $(0,–2,–5).$

Answer 1

Using formula for co-ordinates of centroid of a triangle

The coordinates of the centroid of the triangle, whose vertices are
$(x_1,y_1,z_1),(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ are
$(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3})$

Given two vertices are $A(2,4,6)$ and $B(0,–2,–5)$

Let the third vertex $C(x,y,z)$

$\therefore$ Centroid of $△ABC$,

$G=(\frac{x+2+0}{3},\frac{y+4-2}{3},\frac{z+6-5}{3})$

$\Rightarrow G=(\frac{x+2}{3},\frac{y+2}{3},\frac{z+1}{3})$


Simplifying using given coordinates of centroid

Given centroid is at origin
$\therefore G(\frac{x+2}{3},\frac{y+2}{3},\frac{z+1}{3})=G(0,0,0)$

Comparing coordinates, we get

$\frac{x+2}{3}=0\Rightarrow x=-2$

$\frac{y+2}{3}=0\Rightarrow y=-2$

$\frac{z+1}{3}=0\Rightarrow z=-1$

Therefore, third vertex of the triangle is $(–2,–2,–1)$