Question
Find the centroid of $\triangle\text{ABC}$ whose vertices are A(2, 2), B(-4, -4) and C(5, -8).
✓
Answer
The given points are $A(2,2), B(-4,-4)$ and $C(5,-8)$. Here, $\left( x _1=2, y _1=2\right),\left( x _2=-4, y _2=-4\right)$ and $\left( x _3=5, y _3=-8\right)$
Let G(x, y) be the centroid of $\triangle\text{ABC}.$ Then,
$\text{x}=\frac{1}{3}(\text{x}_1+\text{x}_2+\text{x}_3)$
$=\frac{1}{3}(2-4+5)$
$=1$
$\text{y}=\frac{1}{3}(\text{y}_1+\text{y}_2+\text{y}_3)$
$=\frac{1}{3}(2-4-8)$
$=\frac{-10}{3}$
Hence, the centroid of $\triangle\text{ABC}$ is $\text{G}\Big(1,\frac{-10}{3}\Big).$
Let G(x, y) be the centroid of $\triangle\text{ABC}.$ Then,
$\text{x}=\frac{1}{3}(\text{x}_1+\text{x}_2+\text{x}_3)$
$=\frac{1}{3}(2-4+5)$
$=1$
$\text{y}=\frac{1}{3}(\text{y}_1+\text{y}_2+\text{y}_3)$
$=\frac{1}{3}(2-4-8)$
$=\frac{-10}{3}$
Hence, the centroid of $\triangle\text{ABC}$ is $\text{G}\Big(1,\frac{-10}{3}\Big).$
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