Question

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

Answer 1

Step-by-step explanation:

X(2,3)

Y(3,1)

Z(-1,2)

Centroid(x,y)=(2+3-1/3 , 3+1+2/3)

(4/3, 6/3)

(4/3, 2)

Answer 2

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

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

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

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

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

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

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

$\large{\pink{\bold{\underline{Now:}}}} \\ \\ \rightarrow \: \sf \: Centroid = ( \frac{2 + 3 + (-1)}{3}) ,( \frac{3 + 1 + 2}{3} ) \\ \\ \rightarrow \sf \: Centroid = ( \frac{5 - 1}{3}) , (\frac{\cancel6}{\cancel3} ) \\ \\ \rightarrow \sf \: Centroid = ( \frac{4}{3}) , (\frac{2}{1} ) \\ \\ \rightarrow \sf \: Centroid = (1.3,2)$

$\large{\orange{\bold{\underline{Therefore:}}}} \\ \\ \sf \: The \: coordinates \: of \: Centroid \: of \: the \: triangle \\ \sf \: is \: (1.3,2).$