Question

Find the centroid of a triangle,whose are (6,2),(0,0)and(4,-5)

Answer 1

Answer:

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$x = \frac{x1 + x2 + x3}{3} \\ x = \frac{6 + 0 + 4}{3} \\ x = \frac{10}{3} \\ \\ y = \frac{y1 + y2 + y3}{3} \\ y = \frac{2 + 0 - 5}{3} \\ y = \frac{ - 3}{3} \\ y = - 1$

x = 10/3 and y = -1

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Answer 2

Given:

• The vertices of the triangle are (6, 2), (0, 0) and (4, -5).

To find:

• The centroid.

Answer:

Centroid formula:

$\tt \Bigg( \dfrac{x_1\ +\ x_2\ +\ x_3}{3},\ \dfrac{y_1\ +\ y_2\ +\ y_3}{3}\Bigg)$

From the data we have,

$\tt x_1\ =\ 6\\\\x_2\ =\ 0\\\\x_3\ =\ 4\\\\y_1\ =\ 2\\\\y_2\ =\ 0\\\\y_3\ =\ -5$

Using them in the formula,

$\tt \Bigg(x,\ y\Bigg)\ =\ \Bigg(\dfrac{6\ +\ 0\ +\ 4}{3},\ \dfrac{2\ +\ 0\ +\ -5}{3}\Bigg)\\\\\\\Bigg(x,\ y\Bigg)\ =\ \Bigg(\dfrac{10}{3},\ \dfrac{-3}{3}\Bigg)$

Therefore,

the centroid of the triangle is (10/3, -1).