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Question 3 (iii)
The points A(x1,y1),B(x2,y2) and C(x3,y3) are the vertices of Δ 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.


Solution

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


⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢Since, BE is the median of side CA, so BE divides AC into two equal partsmidpoint of AC=coordinate of E E=(x1+x32,y1+y32)⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢Since, CF is the median of side AB, so CF divides AB into two equal partsmidpoint of AB=coordinate of F F=(x1+x22,y1+y22)⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

Given:
The point Q(p,q), divide the line joining
B(x2,y2) and E(x1+x32,y1+y32) (as E is the mid point of AC) in the ratio 2:1.
and 
The point R(r,s), divide the line joining C(x3,y3) and F(x1+x22,y1+y22) (as F is the mid point of AB) in the ratio 2:1.

The coordinates of Q

Q2.(x1+x32+1.x2)2+1,2.(y1+y32+1.y2)2+1=Q(x1+x2+x33,y1+y2+y33)

The coordinate of R

R2.(x1+x22+1.x3)2+1,2.(y1+y22+1.y3)2+1=R(x1+x2+x33,y1+y2+y33)

So, the required coordinate of points R and Q=(x1+x2+x33,y1+y2+y33)

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

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