#### (i) The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of $\triangle$ABC.The median from A meets BC at D. Find the coordinates of the point D. (ii) The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of $\triangle$ABC.Find the coordinates of the point P on AD such that AP : PD = 2 : 1 (iii) The points A(x1, y1)B(x2, y2)and C(x3, y3)are the vertices of $\triangle$ABC.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 (iv) The points A(x1, y1),B(x2, y2) and C(x3, y3) are the vertices of $\triangle$ABC.What are the coordinates of the centroid of the triangle ABC?

(i) Solution

D is the mid-point of BC.
$Mid-point \,formula = \left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$

Coordinates of $D\left ( x,y \right )= \left ( \frac{x_{2}+x_{3}}{2} ,\frac{y_{2}+y_{3}}{2}\right )$ (By midpoint formula)
(ii) Solution

$Section \, formula= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$D= \left ( \frac{x_{2}+x_{3}}{2},\frac{y_{2}+y_{3}}{2} \right )$
(By Midpoint formula)
$P= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$P= \left ( \frac{2\times \frac{\left ( x_{2}+x_{3} \right )}{2}+1\times x_{1}}{2+1}, \frac{2\times \frac{\left ( y_{2}+y_{3} \right )}{2}+1\times x_{1}}{2+1} \right )$
$P= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$
(iii) Solution

E is mid-point of AC
$E = \left ( \frac{x_{1}+x_{3}}{2} ,\frac{y_{1}+y_{3}}{2}\right )$
Q divides BF at 2 : 1
$Q= \left ( \frac{2\times \frac{\left ( x_{1}+x_{3} \right )}{2}+1\times x_{2}}{2+1}, \frac{2\times \frac{\left ( y_{1}+y_{3} \right )}{2}+1\times y_{2}}{2+1} \right )$
$Q= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$

R divides CF at 2 : 1
$R= \left ( \frac{2\times \frac{\left ( x_{1}+x_{2} \right )}{2}+1\times x_{3}}{2+1}, \frac{2\times \frac{\left ( y_{1}+y_{2} \right )}{2}+1\times y_{3}}{2+1} \right )$
$R= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$
(iv) solution

$co-ordinate\, of\,centroid= \left ( \frac{Sum\,of\,all\,coordinates \, of\,all\,vertices\,}{3}, \frac{Sum\,of\,all\,coordinates \, of\,all\,vertices\,}{3}\right )$
Centroid:
The centroid is the centre point of the triangle which is the intersection of the medians of a triangle.
$\triangle ABC\, coordinates\, of\, centroid= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$