Show that the coordinates of the centroid of the triangle with vertices $A (x_{1}, y_{1}, z_{1})$ , $B (x_{2}, y_{2}, z_{2})$ and $C (x_{3}, y_{3}, z_{3})$ is $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$ .

Open in App

Solution

Let D be the mid - point of BC.

Then, coordinates of D are

$(\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}, \frac{z_{2} + z_{3}}{2})$

Let G be the centroid of $Δ A B C$ , then G divides AD in the ratio 2 : 1.

So, coordinates of G are

$⎡ ⎢ ⎣ \frac{1. x_{1} + 2 (\frac{x_{2} + x_{3}}{2})}{1 + 2}, \frac{1. y_{1} + 2 (\frac{y_{2} + y_{3}}{2})}{1 + 2}, \frac{1. z_{1} + 2 (\frac{z_{2} + z_{3}}{2})}{1 + 2} ⎤ ⎥ ⎦$

i. e. $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$

Hence proved.

Suggest Corrections

Similar questions

Show that the centroid of the triangle with vertices A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3) is
((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3).

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

The coordinates of the mid-point of the line segment joining two points P(X₁, y₁, z₁) and Q(x₂, y₂, z₂) are:

A (x_{1}, y_{1}), B x_{2}, y_{2}), C (x_{3}, y_{3})

(x_{1} - 2)^{2} + (y_{1} - 3)^{2} = (x_{2} - 2)^{2} + (y_{2} - 3)^{2} = (x_{3} - 2)^{2} + (y_{3} - 3)^{2}

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .