Question

find the centroid of a triangle pqr whose vertices are p(1,1)Q(2,2)R(-3,-3)​

Answer 1

Centroid of a triangle= 1+2+(-3)/3,1+2+(-3)/3= (0,0).

Check ur answer.

Answer 2

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

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

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

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

• According to given question :

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