Question
Find the centroid of a $\triangle\text{ABC}$ whose vertices are $A(-1, 0), B(5, -2)$ and $C(8, 2)$
✓
Answer
Here $(x_1 = -1, y_1 = 0)(x_2 = 5, y_2 = -2)$ and $(x_3 = 8, y_3 = 2)$
Let G(x, y) be the centroid of $\triangle\text{ABC},$ then
$\text{x}=\frac{1}{3}(\text{x}_1+\text{x}_2+\text{x}_3)$
$=\frac{1}{3}(-1+8+5)=4$
$\text{y}=\frac{1}{3}(\text{y}_1+\text{y}_2+\text{y}_3)$
$=\frac{1}{3}(0-2+2)=0$
Hence, the centroid of $\triangle\text{ABC},$ is $G(4, 0)$.
Let G(x, y) be the centroid of $\triangle\text{ABC},$ then
$\text{x}=\frac{1}{3}(\text{x}_1+\text{x}_2+\text{x}_3)$
$=\frac{1}{3}(-1+8+5)=4$
$\text{y}=\frac{1}{3}(\text{y}_1+\text{y}_2+\text{y}_3)$
$=\frac{1}{3}(0-2+2)=0$
Hence, the centroid of $\triangle\text{ABC},$ is $G(4, 0)$.
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