Question

Question 7 (iii)
Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of triangle ABC.
(c) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR:RF = 2:1.

Answer 1

Coordinates of E can be calculated as follows:
$x=\frac{1+4}{2}=\frac{5}{2}$
$y=\frac{4+2}{2}=3$
Point Q and P would be coincident because medians of a triangle intersect each other at a common point called centroid.
Median BE of the triangle will divide the side AC in two equal parts.
Therefore, coordinate of Q can be given as follows:
$x=\frac{1\times 6+2\times \frac{5}{2}}{3}=\frac{11}{3}$
$y=\frac{1\times 5+2\times 3}{3}=\frac{11}{3}$

Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid point of side AB.
Coordinates of F = $(\frac{4+6}{2},\frac{2+5}{2})=(5,\frac{7}{2})$

Point R divides the side CF in a ratio 2:1
Coordinates of R = $(\frac{2\times 5+1\times 1}{2+1},\frac{2\times \frac{7}{2}+1\times 4}{2+1})=(\frac{11}{3},\frac{11}{3})$