Question

The points $A$ $(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are the vertices of $\Delta$ ABC.
The median $AD$ meets $BC$ at $D$.

Find the coordinates of points Q and R on medians BE and CF, respectively such that $BQ:QE=2:1$ and $CR:RF=2:1$.

• A. $\frac{2x_1+2x_2+x_3}{3},\frac{2y_1+2y_2+y_3}{3}$
• B. $\frac{x_1+x_2+2x_3}{3},\frac{y_1+y_2+2y_3}{3}$
• C. $x_1+x_2+x_{3}3,y_1+y_2+y_{3}3$
• D. $\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}$

Answer 1

Answer: (D)

$\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}$

The points $A(x_1,y_1),B(x_2,y_2)$ and $C\left(x_3{y}_3\right)$ are the vertices of $\Delta ABC$.

$\therefore$ the median $BE$ from $B$ will meet $AC$ at $E$ which is the mid point of $AC$.

So, by section formula for mid point, the co-ordinates of $E$ is $E(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})=(x_4,y_4)$ .

$\therefore$ the co-ordinates of $Q(x,y)$ are by the section formula,

$x=\frac{m{x}_1+n{x}_4}{m+n}=\frac{\frac{x_1+x_3}{2}\times 2+x_2\times 1}{2+1}=\frac{x_1+x_2+x_3}{3}$ and $y=\frac{m{y}_1+n{y}_4}{m+n}=\frac{\frac{y_1+y_3}{2}\times 2+y_2\times 1}{2+1}=\frac{y_1+y_2+y_3}{3}$

$\therefore Q(x,y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$.

Proceeding in the same way the co-ordinates of $R(p,q),$ which divides the median $CF$ in the ratio $2:1$, are

$p=\frac{\frac{x_1+x_2}{2}\times 2+x_3\times 1}{2+1}=\frac{x_1+x_2+x_3}{3}$ and

$q=\frac{\frac{y_1+y_2}{2}\times 2+y_3\times 1}{2+1}=\frac{y_1+y_2+y_3}{3}$

$\therefore R(p,q)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

So, $Q$ and $R$ are same points.

$Q(x,y)=R(p,q)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$