Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

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Solution

Draw the median AD. D is the midpoint of BC. So D is $\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}$ .

Let G be the centroid. It divides AD in the ratio 2:1. So using the section formula with m = 2 and n =1

This is an important result to remember. Observe that the formula for the centroid is symmetric with respect to $x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3}$ so it doesn't matter in what order the vertices are taken.

$\frac{(x_{1} + x_{2} + x_{3})}{3}, \frac{(y_{1} + y_{2} + y_{3})}{3}$ .

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Q. Question 3 (iv)
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Find the centroid of the triangle with vertices A(x1,y1), B(x2,y2) and C(x3,y3). __

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

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