Question

find the centroid of a triangle ABC whose vertices are (- 1,0 ),(5,- 2) and( 8,2)​

Answer 1

Answer:

centroid of ∆=

$x = \frac{ {x}^{1} + {x}^{2} + {x}^{3} }{3} \\ = \frac{ - 1 + 5 + 8}{3} = \frac{12}{3} = 4 \\ \bold{\underline{x = 4} }$

$y = \frac{ {x}^{1} + {x}^{2} + {x}^{3} }{3} \\ = \frac{0 - 2 + 2}{3} = 0 \\ \bold{\underline{y = 0}}$

hope it helps u.....