If A(−1, 0), B(5, −2) and C(8, 2) are the vertices of a ∆ABC, then its centroid is

(d) (4, 0)

The given points are $A(-1,0), B(5,-2)$ and $C(8,2)$.

Here, $\left(x_{1}=-1, y_{1}=0\right),\left(x_{2}=5, y_{2}=-2\right)$ and $\left(x_{3}=8, y_{3}=2\right)$

Let $G(x, y)$ be the centroid of $\Delta A B C$. Then,

$x=\frac{1}{3}\left(x_{1}+x_{2}+x_{3}\right)$

$=\frac{1}{3}(-1+5+8)$

$=4$

and

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

$=\frac{1}{3}(0-2+2)$

$=0$

Hence, the centroid of $\Delta A B C$ is $G(4,0)$.