Question

find the centroid of the triangle whose vertices are (0,1), (-1,2) and (-2,-3)

Answer 1

x1+x2+x3/3 , y1+y2+y3/3

;0+(-1)+(-2)/3 , 1+2+(-3)/3

;-1-2/3 , 3-3/3

;-3/3 ,0/3

point is (-1,0)

is your answer

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Centroid(G)=}(-1,0)}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ : \implies \text{Coordinate \: of \: A = (0,1)} \\ \\ : \implies \text{Coordinate \: of \: B = (-1,2)} \\ \\ : \implies \text{Coordinate \: of \: C = (-2,-3)} \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \text{Centroid(G) = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \circ \: \text{Centroid \: of \: triangle(G}) \\ \\ \circ \: \text{For \: x }= \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \circ \: \text{For \: y} = \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ \text{Let \: Coordinate \: of \: (g) =( x,y) } \\ \\ \bold{For \: x}\\ : \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ : \implies x = \frac{0+(-1) + (-2)}{3} \\ \\ : \implies x = \frac{-3}{3} \\ \\ \green{: \implies x =-1} \\ \\ \bold{For \: y}\\ : \implies y= \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ : \implies y= \frac{1 +2+(-3)}{3} \\ \\ : \implies y = \frac{0}{3} \\ \\ \green{: \implies y =0} \\ \\ \green{\therefore \text{Coordinate \: of \: centroid(G) = }(-1,0)}$