Let $P(x, 3)$ divide the line segment joining the points $A(-4,2)$ and $B(3,6)$ in the ratio $k: 1$ $\therefore$ Coordinates of $P$ is $\left( \frac{m_1 x_2+m_2 x_1}{m_1+m_2},\; \frac{m_1 y_2+m_2 y_1}{m_1+m_2} \right)$ $= \left( \frac{3 k-4}{k+1},\; \frac{6 k+2}{k+1} \right)$But coordinate of $P$ is $(x, 3)$ $\Rightarrow \;\; \frac{6k+2}{k+1}=3$ $6k+2=3k+3$ $3k=1 \Rightarrow k=\frac{1}{3}$$\therefore$ The required ratio is $\frac{1}{3}: 1$ i.e., $1: 3$ (internally) (i) Now $x=\frac{3k-4}{k+1}$Putting $k=\frac{1}{3}$ , we get $x=\frac{3 \times \frac{1}{3} - 4}{\frac{1}{3} + 1}$$= \frac{1-4}{\frac{1+3}{3}} = \frac{-3}{\frac{4}{3}} = \frac{-9}{4}$ (ii) $\therefore$ Coordinate of $P$ is $\left( \frac{-9}{4}, 3 \right)$ $\text{Length of AP} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ $= \sqrt{ \left( -\frac{9}{4} + 4 \right)^2 + (3-2)^2 }.$ $= \sqrt{ \left( \frac{-9+16}{4} \right)^2 + (1)^2 }$ $= \sqrt{ \frac{49}{16} + 1 } = \sqrt{ \frac{49+16}{16} }$$= \sqrt{ \frac{65}{16} } = \frac{ \sqrt{65} }{4}$