Question

The points $A\left(x_1,y_1\right),B\left(x_2,y_2\right),C\left(x_3,y_3\right)$ are the vertices of $∆$ ABC .
(i) The median from A meets BC at D . Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) 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.
(iv) What are the coordinates of the centropid of the triangle ABC ?

Answer 1

(i) Median AD of the triangle will divide the side BC in two equal parts.



Therefore, D is the midpoint of side BC.
Coordinates of D are
$\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$

(ii)
THe point P divided the side AD in the ratio 2: 1.
Coordinates of P are
$\left(\frac{2\times {\displaystyle \left(\frac{x_2+x_3}{2}\right)}+1\times x_1}{2+1},\frac{{\displaystyle 2\times \left(\frac{y_2+y_3}{2}\right)+1\times y_1}}{2+1}\right)=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$


(iii)
Median BE of the triangle will divide the side AC in two equal parts.
Therefore, E is the midpoint of side AC.
Coordinates of E are
$\left(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\right)$
The point Q divided the side BE in the ratio 2: 1.
Coordinates of Q are
$\left(\frac{2\times {\displaystyle \left(\frac{x_1+x_3}{2}\right)}+1\times x_2}{2+1},\frac{{\displaystyle 2\times \left(\frac{y_1+y_3}{2}\right)+1\times y_2}}{2+1}\right)=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

Similarly, Coordinates of Q are R are $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$.

(iv)
The points P, Q and R coincides and is the centroid of the triangle ABC.
So, coordinates of the centroid is $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$.