Question

the coordinates of the centre of centroid of the triangle whose vertices are (0,6) (8,12) (8,0)​

Answer 1

$\rm\large\underline{\text{Theorem}}$

The centroid of the circle is $\textrm{G (\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}) }$, provided that three vertices of triangles are $(x_i,y_i)(i=1,2,3)$.

$\rm\large\underline{\text{Solution}}$

Let the three vertices be $\textrm{A(0,6), B(8,12), C(8,0)}$.

And let the centroid of the circle be $\textrm{G (x,y) }$.

We know that,

$\cdots\longrightarrow\textrm{G (\dfrac{0+8+8}{3},\dfrac{6+12+0}{3}) }$

$\cdots\longrightarrow\textrm{G (\dfrac{16}{3},6) }$

The centroid of the triangle is $\rm G(\dfrac{16}{3},6)$.

$\rm\large\underline{\text{Proof}}$

(Assertion)

The three medians of a triangle divide each other internally in the ratio $2:1$.

(Result)

Consider the midpoint $(\dfrac{x_1+x_2}{2})$ which is the midpoint of $(x_1,y_1)$ and $(x_2,y_2)$.

Two points $(x_3,y_3)$ and $(\dfrac{x_1+x_2}{2})$ divide each other in the ratio of $2:1$.

By internal section formula,

$\cdots\longrightarrow\text{G (\dfrac{2\times\dfrac{x_1+x_2}{2}+x_3}{2+1} , \dfrac{2\times\dfrac{y_1+y_2}{2}+y_3}{2+1}) }$

$\cdots\longrightarrow\text{G (\dfrac{x_1+x_2+x_3}{3} , \dfrac{y_1+y_2+y_3}{3}) }$

Hence proven.