Find the centroid of ∆ABC whose vertices are A(−1, 0), B(5, −2) and C(8, 2).

Here, (x1 = −1, y1 = 0), (x2 = 5, y2 = −2) and (x3 = 8, y3 = 2).
Let G(x, y) be the centroid of the ∆ABC. Then,

$x=\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right)=\frac{1}{3}(-1+5+8)=\frac{1}{3}(12)=4$

$y=\frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right)=\frac{1}{3}(0-2+2)=\frac{1}{3}(0)=0$

Hence, the centroid of ∆ABC is G(4, 0).