The points $A (x_1, y_1), B (x_2, y_2)$ and $C (x_3, y_3)$ are the vertices of $\triangle ABC$.
Find the points of coordinates $Q$ and $R$ on medians $BE$ and $CF$ respectively such that $BQ : QE = 2 : 1$ and $CR : RF = 2 : 1$.

Given:

The points $A (x_1, y_1), B (x_2, y_2)$ and $C (x_3, y_3)$ 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 $D=(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2})$.

$AP : PD = 2 : 1$.

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 \( (\frac{2 \times \frac{x_2+x_3}{2}+1 \times x_1}{1+2}, \frac{2 \times \frac{y_2+y_3}{2}+1 \times y_1}{1+2}) \)

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

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

Similarly,

Coordinates of $E=(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2})$.

The coordinates of \( \mathrm{Q} \) are \( (\frac{2 \times \frac{x_1+x_3}{2}+1 \times x_2}{1+2}, \frac{2 \times \frac{y_1+y_3}{2}+1 \times y_2}{1+2}) \)

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

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

Coordinates of $F=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.

$CR : RF = 2 : 1$

The coordinates of \( \mathrm{R} \) are \( (\frac{2 \times \frac{x_1+x_2}{2}+1 \times x_3}{1+2}, \frac{2 \times \frac{y_1+y_2}{2}+1 \times y_3}{1+2}) \)

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

Tutorialspoint

Updated on: 10-Oct-2022

34 Views

Kickstart Your Career

Get certified by completing the course

Get Started