Calculate the ratio in which the line joining $A \left(-4, 2\right)$ and $B \left(3, 6\right)$ is divided by the point $P \left(x, 3\right)$.
Also find the coordinates of point $P$ and the length of $AP$.
Given: $\;$ $A \left(-4, 2\right)$, $\;$ $B \left(3, 6\right)$
Let the point $P \left(x,3\right)$ divide the line joining the points $A$ and $B$ in the ratio $k : 1$
Then, by section formula [for $Y$ coordinate of point $P$]
$\dfrac{6k + 2 \times 1}{k + 1} = 3$
i.e. $\;$ $6k + 2 = 3k +3$
i.e. $\;$ $3k = 1$ $\implies$ $k = \dfrac{1}{3}$
i.e. $\;$ the point $P$ divides $AB$ in the ratio $1 : 3$
Again, by section formula [for $X$ coordinate of point $P$]
$\dfrac{-4 \times 3 + 3 \times 1}{1 + 3} = x$
i.e. $\;$ $x = \dfrac{-12 + 3}{4} = \dfrac{-9}{4}$
$\therefore \;$ Point $\;$ $P = \left(x, 3\right) = \left(\dfrac{-9}{4}, 3\right)$
By distance formula,
$\begin{aligned}
\text{length of AP} & = \sqrt{\left[-4 - \left(\dfrac{-9}{4}\right)\right]^2 + \left[2 - 3\right]^2} \\\\
& = \sqrt{\left(\dfrac{-7}{4}\right)^2 + \left(-1\right)^2} \\\\
& = \sqrt{\dfrac{49}{16} + 1} \\\\
& = \dfrac{\sqrt{65}}{4}
\end{aligned}$