The points A `(x_(1),y_(1)), B (x_(2),y_(2)) and C (x_(3),y_(3))` are the vertices of `triangle`ABC.
(i) The median from A Meets Bc at D. Find the coordinates of the points D.
(ii) Find the coordinates of the point P on Ad such that `AP : PD = 2 : 1`.
(iii) Find the coordinates of points Q and R on medians BE and CF, respectively such that `BQ : QE=2 :1 and CR : RF =2 : 1`.
What are the coordinates of the centroid of the `triangle` ABC?

Explanation

Given that , the points A `(x_(1),y_(1)), B (x_(2),y_(2)) and C (x_(3),y_(3))` are the vertices of `triangle`ABC.
(i) We know that , the median bisect the line segment into two equal i.e., here D is the mid- point of BC.
`:.` Coordinate of mid - point of BC `=((x_(2)+x_(3))/(2),(y_(2)+y_(3))/(2))`
`rArrD-=((x_(2)+x_(3))/(2),(y_(2)+y_(3))/(2))`
(ii) Let the coordinates of a point P be (x,y).
Given that point P(x,y), divide the line joining A`rArrD((x_(2)+x_(3))/(2),(y_(2)+y_(3))/(2))` in the ratio `2 : 1` , then the coordinates of P

`[(2*((x_(2)+x_(3))/(2))+1*x_(1))/(2+1),(2*((y_(2)+y_(3))/(2)+1*y_(1)))/(2+1)]`
[`:'` internal section formula `=((m_(1)x_(2)+m_(2)x_(1))/(m_(1)+m_(2)),(m_(1)y_(2)+m_(2)y_(1))/(m_(1)+m_(2))]`
`-=((x_(2)+x_(3)+x_(1))/(3),(y_(1)+y_(2)+y_(3))/(2))`
`:.` So , required coordinates of point P `-=((x_(1)+x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))`
(iii) Let the coordinates of a point Q be (pq)

Given that , the point Q (p,q) , divide the line joining B `(x_(2),y_(2))` and E `((x_(1)+x_(3))/(2),(y_(1)+y_(3))/(2))` in the ratio `2 : 1` , then the coordinates of Q
`=[(2.((x_(1)+x_(3))/(2))+1*x_(2))/(2+1),(2.((y_(1)+y_(2))/(2))+1*y_(2))/(2+1)]`
`((x_(1)+x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))`
[ since, BE is the median of side CA, so BE divedes AC in to two equal parts.
`therefore` mid - point of AC= Coordinate of `ErArrE=((x_(1)_x_(3))/(2),(y_(1)+y_(2))/(2))`
So, the required coordinate of point `Q-=((x_(1)+x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))`
Now, let the coordinates of a point E be `(alpha, beta)`. Given that, the point `R(alpha,beta)`, divide the line joining `C(x_(3)y_(3))` and `F((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))` in the ratio `2:1`, then the coordinates of R
`-=[(2.((x_(1)+x_(3))/(2))+1*x_(3))/(2+1),(2.((y_(1)+y_(2))/(2))+1*y_(3))/(2+1)]`
`=((x_(1)+x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))`
[since , CF is the median of side AB. So , CF divides AB in to two equal parts. `:.` mid - point of AB = coordinate of `FrArr((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))]`
So, the required coordinate of point `-=((x_(1)+x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))`
(iv) Coordinates of the centroid of the `triangle` ABC
`((" Sum of abscissa of all vertices")/(3),("Sum of ordinate of all vertices")/(3))`
`=((x_(1)+x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))`