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$A (4, 2), B (6, 5)$ and $C (1, 4)$ are the vertices of $\triangle ABC$.Find the coordinates of the points Q and R on medians BE and CF respectively such that $BQ : QE = 2 : 1$ and $CR : RF = 2 : 1$.What do you observe?
Given:
$A (4, 2), B (6, 5)$ and $C (1, 4)$ are the vertices of $\triangle ABC$.
$BQ : QE = 2 : 1$ and $CR : RF = 2 : 1$.
To do:
We have to find the coordinates of the points Q and R on medians BE and CF respectively.
Solution:
$D$ is the mid-point of BC.
This implies,
Using mid-point formula, we get,
Coordinates of $D=(\frac{6+1}{2}, \frac{5+4}{2})$
$=(\frac{7}{2},\frac{9}{2})$
Similarly,
Coordinates of $E=(\frac{4+1}{2}, \frac{2+4}{2})$
$=(\frac{5}{2},3)$
Coordinates of $F=(\frac{4+6}{2}, \frac{2+5}{2})$
$=(5,\frac{7}{2})$
$AP : PD = 2 : 1$.
Using section formula, we get,
\( (x,y)=(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}) \)
The coordinates of \( \mathrm{P} \) are \( (\frac{2 \times \frac{7}{2}+1 \times 4}{1+2}, \frac{2 \times \frac{9}{2}+1 \times 2}{1+2}) \)
\( =(\frac{7+4}{3}, \frac{9+2}{3}) \)
\( =(\frac{11}{3}, \frac{11}{3}) \)
$BQ : QE = 2 : 1$The coordinates of \( \mathrm{Q} \) are \( (\frac{2 \times \frac{5}{2}+1 \times 6}{1+2}, \frac{2 \times 3+1 \times 5}{1+2}) \)
\( =(\frac{5+6}{3}, \frac{6+5}{3}) \)
\( =(\frac{11}{3}, \frac{11}{3}) \)
$CR : RF = 2 : 1$
The coordinates of \( \mathrm{R} \) are \( (\frac{2 \times 5+1 \times 1}{1+2}, \frac{2 \times \frac{7}{2}+1 \times 4}{1+2}) \)
\( =(\frac{10+1}{3}, \frac{7+4}{3}) \)
\( =(\frac{11}{3}, \frac{11}{3}) \)
We observe that the coordinates of $P, Q$ and $R$ are the same. $P, Q$ and $R$ coincide with each other.
Medians of the sides of a triangle pass through the same point which is called the centroid of the triangle. Therefore, $(\frac{11}{3},\frac{11}{3})$ is the centroid of the triangle ABC.
The coordinates of \( \mathrm{Q} \) and \( \mathrm{R} \) are \( (\frac{11}{3}, \frac{11}{3}) \).
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