A (4, 2), B(6, 5) and C (1, 4) are the vertices of ΔABC.

Solution:

We have triangle $\triangle \mathrm{ABC}$ in which the co-ordinates of the vertices are $\mathrm{A}(4,2) ; \mathrm{B}(6,5)$ and $\mathrm{C}(1,4)$

(i)It is given that median from vertex A meets BC at D. So, D is the mid-point of side BC.

In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Therefore mid-point D of side BC can be written as,

$\mathrm{D}(x, y)=\left(\frac{6+1}{2}, \frac{5+4}{2}\right)$

Now equate the individual terms to get,

$x=\frac{7}{2}$

$y=\frac{9}{2}$

So co-ordinates of $D$ is $\left(\frac{7}{2}, \frac{9}{2}\right)$

(ii)We have to find the co-ordinates of a point P which divides AD in the ratio 2: 1 internally.

Now according to the section formula if any point $\mathrm{P}$ divides a line segment joining $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ in the ratio $\mathrm{m}$ : $\mathrm{n}$ internally than,

$\mathrm{P}(x, y)=\left(\frac{m x_{1}+m x_{2}}{m+n}, \frac{m y_{1}+m y_{2}}{m+n}\right)$

P divides AD in the ratio 2: 1. So,

$\mathrm{P}(x, y)=\left(\frac{2\left(\frac{7}{2}\right)+4(1)}{1+2}, \frac{2\left(\frac{9}{2}\right)+1(2)}{1+2}\right)$

$=\left(\frac{11}{3}, \frac{11}{3}\right)$

(iii)We need to find the mid-point of sides AB and AC. Let the mid-points be F and E for the sides AB and AC respectively.

Therefore mid-point F of side AB can be written as,

$\mathrm{F}(x, y)=\left(\frac{6+4}{2}, \frac{5+2}{2}\right)$

So co-ordinates of $\mathrm{F}$ is $\left(5, \frac{7}{2}\right)$

Similarly mid-point E of side AC can be written as,

$\mathrm{E}(x, y)=\left(\frac{1+4}{2}, \frac{4+2}{2}\right)$

So co-ordinates of $E$ is $\left(\frac{5}{2}, 3\right)$

Q divides BE in the ratio 2: 1. So,

$\mathrm{Q}(x, y)=\left(\frac{2\left(\frac{5}{2}\right)+6(1)}{1+2}, \frac{2(3)+1(5)}{1+2}\right)$

$=\left(\frac{11}{3}, \frac{11}{3}\right)$

Similarly, R divides CF in the ratio 2: 1. So,

$\mathrm{R}(x, y)=\left(\frac{2(5)+1(1)}{1+2}, \frac{2\left(\frac{7}{2}\right)+1(4)}{1+2}\right)$

$=\left(\frac{11}{3}, \frac{11}{3}\right)$

(iv)We observe that that the point P, Q and R coincides with the centroid. This also shows that centroid divides the median in the ratio 2: 1.