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Co-Ordinate Geometry — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsCo-Ordinate Geometry4 Marks
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, $(x_1 = 2, y_1 = 2), (x_2 = -4, y_2 = -4)$ and $(x_3 = 5, y_3 = -8)$
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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