Question

Find the centroid of triangle whose vertices are 3, 4 - 1, - 2 and 10 - 5

Answer 1

Answer:

centroid=(4,-1)

Step-by-step explanation:

centroid=x,y

x=x1+x2+x3/3

y=y1+y2+y3/3

x=3+-1+10/3

 =4

y=4+-2+-5/3

 =-1

therefore centroid=(4,-1)

Answer 2

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

$\green{\therefore{\text{Centroid(G)=}(4,-1)}}$

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

$\green{ \underline \bold{Given : }} \\ : \implies \text{Coordinate \: of \: A = (3,4)} \\ \\ : \implies \text{Coordinate \: of \: B = (-1,-2)} \\ \\ : \implies \text{Coordinate \: of \: C = (10,-5)} \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \text{Centroid(G) = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \circ \: \text{Centroid \: of \: triangle(G}) \\ \\ \circ \: \text{For \: x }= \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \circ \: \text{For \: y} = \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ \text{Let \: Coordinate \: of \: (g) =( x,y) } \\ \\ \bold{For \: x}\\ : \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ : \implies x = \frac{3+(-1) + 10}{3} \\ \\ : \implies x = \frac{12}{3} \\ \\ \green{: \implies x =4} \\ \\ \bold{For \: y}\\ : \implies y= \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ : \implies y= \frac{4+(-2)+(-5)}{3} \\ \\ : \implies y = \frac{-3}{3} \\ \\ \green{: \implies y =-1} \\ \\ \green{\therefore \text{Coordinate \: of \: centroid(G) = }(4,-1)}$