Question

Find the centroid of a triangle whose vertices are $(2,1),(5,2)$ and $(3,4)$.

• A. $(\frac{10}{3},\frac{7}{3})$
• B. $(\frac{5}{3},\frac{7}{3})$
• C. $(\frac{10}{3},\frac{5}{3})$
• D. $(\frac{5}{3},\frac{5}{3})$

Answer 1

The correct option is A $(\frac{10}{3},\frac{7}{3})$

The centroid of the triangle ABC = $[\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}]$

Let the points be A(2,1) , B(5,2) and C(3,4)

$\therefore$ Centroid of triangle $=(\frac{2+5+3}{3},\frac{1+2+4}{3})=(10/3,7/3)$