Question

Find the centroid of the triangle with vertices A$(x_1,y_1)$, B$(x_2,y_2)$ and C$(x_3,y_3)$.

• A. (x₁+ x₂+ x_­3)/3 , (y₁+ y₂+ y₃)/3
• B. - (x₁+ x₂+ x_­3)/3 , - (y₁+ y₂+ y₃)/3
• C. (x₁ - x₂ - x_­3)/3 , (y₁- y₂ - y₃)/3
• D. (x₁- x₂+ x_­3)/3 , (y₁- y₂+ y₃)/3

Answer 1

The correct option is A

(x₁+ x₂+ x_­3)/3 , (y₁+ y₂+ y₃)/3

Draw the median AD. D is the midpoint of BC. So D is $\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}$.

Let G be the centroid. It divides AD in the ratio 2:1. So using the section formula with m = 2 and n =1

This is an important result to remember. Observe that the formula for the centroid is symmetric with respect to $x_1,x_2,x_3,y_1,y_2,y_3$ so it doesn't matter in what order the vertices are taken.

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