Question

(4,7), (8,4), (7,11) Find the centroids of the triangles whose vertices are given

Answer 1

Let the Points (4,7), (8,4),  and (7,11) be P(x₁, y₁), Q(x₂,y₂), and R(x₃,y₃) respectively. 

Now, Using the Formula,

G(x,y) = [ (x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3 ]
    = [(4 + 8 + 7)/3 , (7 + 4 + 11)/3]
    = [19/3 , 22/3]
    = (19/3, 22/3)



Hence, the co-ordinates of the Point G is (19/3, 22/3).


Hope it helps. 

Answer 2

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

$\green{\therefore{\text{Centroid(G)=}(\frac{19}{3},\frac{22}{3})}}$

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

$\green{ \underline \bold{Given : }} \\ : \implies \text{Coordinate \: of \: A = (4,7)} \\ \\ : \implies \text{Coordinate \: of \: B = (8,4)} \\ \\ : \implies \text{Coordinate \: of \: C = (7,11)} \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \text{Centroid(G) = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \circ \: \text{Centroid \: of \: triangle(G}) \\ \\ \circ \: \text{For \: x }= \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \circ \: \text{For \: y} = \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ \text{Let \: Coordinate \: of \: (g) =( x,y) } \\ \\ \bold{For \: x}\\ : \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ : \implies x = \frac{ 4+ 8 + 7}{3} \\ \\ : \implies x = \frac{19}{3} \\ \\ \green{: \implies x =\frac{19}{3}} \\ \\ \bold{For \: y}\\ : \implies y= \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ : \implies y= \frac{ 7 +4+11}{3} \\ \\ : \implies y = \frac{22}{3} \\ \\ \green{: \implies y =\frac{22}{3}} \\ \\ \green{\therefore \text{Coordinate \: of \: centroid(G) = }(\frac{19}{3},\frac{22}{3})}$