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

1.	Find the centroid of the triangle whose vertices are (0,1), (-1,2) and (-2,-3)
Answer» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\therefore{\text{Centroid(G)=}(-1,0)}}\\$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ $\green{ \underline \bold{<klux>GIVEN</klux> : }} \\ : \implies \text{Coordinate \: of \: A = (0,1)} \\ \\ : \implies \text{Coordinate \: of \: B = (-1,2)} \\ \\ : \implies \text{Coordinate \: of \: C = (-2,-3)} \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \text{Centroid(G) = ?}$ • ACCORDING to given QUESTION : $\bold{As \: we \: know \: that} \\ \circ \: \text{Centroid \: of \: triangle(G}) \\ \\ \circ \: \text{For \: x }= \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \circ \: \text{For \: y} = \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ \text{Let \: Coordinate \: of \: (g) =( x,y) } \\ \\ \bold{For \: x}\\ : \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ : \implies x = \frac{0+(-1) + (-2)}{3} \\ \\ : \implies x = \frac{-3}{3} \\ \\ \green{: \implies x =-1} \\ \\ \bold{For \: y}\\ : \implies y= \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ : \implies y= \frac{1 +2+(-3)}{3} \\ \\ : \implies y = \frac{0}{3} \\ \\ \green{: \implies y =0} \\ \\ \green{\therefore \text{Coordinate \: of \: centroid(G) = }(-1,0)}$

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

Answer»

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

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

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

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

• ACCORDING to given QUESTION :

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

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

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