Question

Question 3 (iii)
The points $A(x_1,y_1),B(x_2,y_2)andC(x_3,y_3)$ are the vertices of $\Delta$ ABC.
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.

Answer 1

Let the coordinates of the points Q be (p,q) and R be (r,s)


$$\left[\begin{array}{c}\\ Since,BEisthemedianofsideCA,\\ soBEdividesACintotwoequalparts\\ \therefore mid-pointofAC=coordinateofE\\ \Rightarrow E=(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})\end{array}\right]$$

$$\left[\begin{array}{c}\\ Since,CFisthemedianofsideAB,\\ soCFdividesABintotwoequalparts\\ \therefore mid-pointofAB=coordinateofF\\ \Rightarrow F=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\end{array}\right]$$

Given:
The point $Q(p,q)$, divide the line joining $B(x_2,y_2)$ and $E(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})$ (as E is the mid point of AC) in the ratio 2:1.
and
The point $R(r,s)$, divide the line joining $C(x_3,y_3)$ and $F(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ (as F is the mid point of AB) in the ratio 2:1.

The coordinates of Q

$Q\equiv (\frac{2.(\frac{x_1+x_3}{2}+1.x_2)}{2+1},\frac{2.(\frac{y_1+y_3}{2}+1.y_2)}{2+1})=Q(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

The coordinate of R

$R\equiv (\frac{2.(\frac{x_1+x_2}{2}+1.x_3)}{2+1},\frac{2.(\frac{y_1+y_2}{2}+1.y_3)}{2+1})=R(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$So,therequiredcoordinateofpointsRandQ=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Hence the points Q and R are the same point as they have the same co-ordinates.
Here it is the "centroid" as it is the point of intersection of medians