Question

find the coordinates of the centroid of a triangle whose vertices are (0,6),(8,12)&(8,0)?

Answer 1

$centroid \: = (\frac{x1 + x2 + x3}{3} . \frac{y1 + y2 + y3}{3} ) \\ = (\frac{0 + 8 + 8}{3} . \frac{6 + 12 + 0}{3} ) \\ = (\frac{16}{3} . \frac{18}{3} )$
Centroid = (16/3 , 18/3)

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Answer 2

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

$\green{\therefore{\text{Centroid(G)=}(\frac{16}{3},6)}}$

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

$\green{ \underline \bold{Given : }} \\ : \implies \text{Coordinate \: of \: A = (0,6)} \\ \\ : \implies \text{Coordinate \: of \: B = (8,12)} \\ \\ : \implies \text{Coordinate \: of \: C = (8,0)} \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \text{Centroid(G) = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \circ \: \text{Centroid \: of \: triangle(G}) \\ \\ \circ \: \text{For \: x }= \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \circ \: \text{For \: y} = \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ \text{Let \: Coordinate \: of \: (g) =( x,y) } \\ \\ \bold{For \: x}\\ : \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ : \implies x = \frac{0+8 + 8}{3} \\ \\ : \implies x = \frac{8+8}{3} \\ \\ \green{: \implies x =\frac{16}{3}} \\ \\ \bold{For \: y}\\ : \implies y= \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ : \implies y= \frac{6+12+0}{3} \\ \\ : \implies y = \frac{18}{3} \\ \\ \green{: \implies y =6} \\ \\ \green{\therefore \text{Coordinate \: of \: centroid(G) = }(\frac{16}{3},6)}$