Let $\mathrm{P}\left(\mathit{x},3\right)$ divide the line segment joining the points $\mathit{A}\left(-4,2\right)$ and $\mathit{B}\left(3,6\right)$ in the ratio $\mathit{k}:1$

$\therefore$ Coordinates of $\mathrm{P}$ is
$\left(\frac{{\mathit{m}}_1{\mathit{x}}_2+{\mathit{m}}_2{\mathit{x}}_1}{{\mathit{m}}_1+{\mathit{m}}_2},\frac{{\mathit{m}}_1{\mathit{y}}_2+{\mathit{m}}_2{\mathit{y}}_1}{{\mathit{m}}_1+{\mathit{m}}_2}\right)$ $=\left(\frac{3\mathit{k}-4}{\mathit{k}+1},\frac{6\mathit{k}+2}{\mathit{k}+1}\right)$
But coordinate of $\mathrm{P}$ is $\left(\mathit{x},3\right)$
$\Rightarrow \frac{6\mathit{k}+2}{\mathit{k}+1}=3$
$6\mathit{k}+2=3\mathit{k}+3$
$3\mathit{k}=1\Rightarrow \mathit{k}=\frac{1}{3}$
$\therefore$ The required ratio is $\frac{1}{3}:1$ i.e., $1:3$ (internally)
(i) Now $\mathit{x}=\frac{3\mathit{k}-4}{\mathit{k}+1}$
Putting $\mathit{k}=\frac{1}{3}$ , we get
$\mathit{x}=\frac{3\times \frac{1}{3}-4}{\frac{1}{3}+1}$
$=\frac{1-4}{\frac{1+3}{3}}=\frac{-3}{4∕3}=\frac{-9}{4}$

(ii) $\therefore$ Coordinate of $\mathrm{P}$ is $\left(\frac{-9}{4},3\right)$

$=\sqrt{{\left(-\frac{9}{4}+4\right)}^2+{\left(3-2\right)}^2}.$
$=\sqrt{{\left(\frac{-9+16}{4}\right)}^2+{\left(1\right)}^2}$
$=\sqrt{\frac{49}{16}+1}=\sqrt{\frac{49+16}{16}}$
$=\sqrt{\frac{65}{16}}=\frac{\sqrt{65}}{4}$