Question

find the Centroid of the triangle whose vertices are x(4,-1),y(1,2)&z(3,-1).​

Answer 1

Explanation:

$\large{\red{\bold{\underline{Given:}}}}$

$\sf \: Coordinates \: of \: the \: vertices \: are: \: x(4,-1), \: y(1,2) \: and \: z(3,-1)$

$\large{\green{\bold{\underline{To \: Find:}}}}$

$\sf \: Centroid \: of \: the \: triangle$

$\large{\blue{\bold{\underline{Formula \: Used:}}}} \\ \\ \sf \:Coordinates \: of \:Centroid = \: (\frac{x_{1} + x_{2} + x_{3}}{3} ),( \frac{y_{1} + y_{2} + y_{3}}{3})$

$\large{\red{\bold{\underline{Solution:}}}} \\ \\ \: \sf \: On \: considering \: the \: respective \: coordinates \: as :$

$\sf \: x(4,-1) \: \rightarrow \: (x_{1}, y_{1}) \\ \\\sf \: y(1,2) \: \rightarrow \: (x_{2}, y_{2}) \\ \\ \sf \: z(3,-1) \: \rightarrow \: (x_{3}, y_{3})$

$\large{\pink{\bold{\underline{Now:}}}} \\ \\ \rightarrow \: \sf \: Centroid = ( \frac{4 + 1 + 3}{3}) ,( \frac{-1 + 2 + (-1)}{3} ) \\ \\ \rightarrow \sf \: Centroid = ( \frac{5 + 3}{3}) , (\frac{-2 + 2}{3} ) \\ \\ \rightarrow \sf \: Centroid = ( \frac{8}{3} ),( \frac{0}{3} ) \\ \\ \rightarrow \sf \: Centroid = (2.6,0)$

$\large{\orange{\bold{\underline{Therefore:}}}} \\ \\ \sf \: The \: coordinates \: of \: Centroid \: of \: the \: triangle \\ \sf \: is \: (2.6,0).$