Question

8) Find the
co-ordinates of a centroid of triangle whose vertices
are (6,2), (0,0) and ( 4,-7)​

Answer 1

✫ The centroid of triangle is (10/3,-5/3). ✫

Step-by-step explanation:

As we know that,

$\;\;\;\;\;\bullet\;\boldsymbol{G = \bigg(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\bigg)}$

Here,

• The point of concurrence of medians of a triangle is called centroid of triangle,it is denoted by 'G'.
• Vertices = (6,2), (0,0), ( 4,-7)​
• x₁ = 6,y₁ = 2
• x₂ = 0,y₂ = 0
• x₃ = 4,y₃ = -7

Applying the values,

$\\ \dashrightarrow\sf{G = \frac{6+0+4}{3},\frac{2+0+(-7)}{3}}\\ \\ \dashrightarrow\sf{G = \frac{10}{3},\frac{-5}{3}}$

• ∴ The centroid of triangle is (10/3,-5/3).

Answer 2

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

★$\begin{gathered}\green{\therefore{\text{Centroid(G)=}(\frac{10}{3},\frac{-5}{3})}}\\\end{gathered}$★

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

$\begin{gathered} \green{ \underline \bold{Given : }} \\ : \implies \text{Coordinate \: of \: A = (6,2)} \\ \\ : \implies \text{Coordinate \: of \: B = (0,0)} \\ \\ : \implies \text{Coordinate \: of \: C = (4,-7)} \\ \\ \red{ \underline \bold{To \: Find : }} \\ : \implies \text{Centroid(G) = ?}\end{gathered}$

• :⟹Coordinate of A = (6,2)
• :⟹Coordinate of B = (0,0)
• :⟹Coordinate of C = (4,-7)

ToFind:

:⟹Centroid(G) = ?

• According to given question :

$\begin{gathered} \bold{As \: we \: know \: that} \\ \circ \: \text{Centroid \: of \: triangle(G}) \\ \\ \circ \: \text{For \: x }= \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \circ \: \text{For \: y} = \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ \text{Let \: Coordinate \: of \: (g) =( x,y) } \\ \\ \bold{For \: x}\\ : \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ : \implies x = \frac{ 6 + 0 + 4}{3} \\ \\ : \implies x = \frac{6+4}{3} \\ \\ \green{: \implies x =\frac{10}{3}} \\ \\ \bold{For \: y}\\ : \implies y= \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ : \implies y= \frac{ 2 +0+(-7)}{3} \\ \\ : \implies y = \frac{2-7}{3} \\ \\ \green{: \implies y =\frac{-5}{3}} \\ \\ \green{\therefore \text{Coordinate \: of \: centroid(G) = }(\frac{10}{3},\frac{-5}{3})}\end{gathered}$