The line segment joining $A \left(-1, \dfrac{5}{3}\right)$ and $B \left(a, 5\right)$ is divided internally in the ratio $1 : 3$ at $P$, the point where the line segment intersects Y axis. Calculate the value of '$a$' and the coordinates of $P$.
Let $\;$ $A \left(-1, \dfrac{5}{3}\right) \equiv \left(x_1, y_1\right)$; $\;$ $B \left(a, 5\right) \equiv \left(x_2, y_2\right)$
Let $P$ divide the line segment $AB$ internally in the ratio $m:n$ $\;$ i.e. $\;$ $1 : 3$
Let the coordinates of point $P$ be $\left(0,y\right)$ $\;\;$ [$\because \; P \;$ lies on the Y axis]
Then, by section formula, $\;$ $0 = \dfrac{m x_1 + n x_2}{m + n}$
i.e. $\;$ $0 = \dfrac{1 \times \left(-1\right) + 3 \times a}{1 + 3}$
i.e. $\;$ $-1 + 3a = 0$ $\implies$ $a = \dfrac{1}{3}$
and, $\;$ $y = \dfrac{m y_1 + n y_2}{m + n}$
i.e. $\;$ $y = \dfrac{1 \times \dfrac{5}{3} + 3 \times 5}{1 + 3}$
i.e. $\;$ $y = \dfrac{\dfrac{5}{3} + 15}{4} = \dfrac{50}{12} = \dfrac{25}{6}$
$\therefore \;$ The coordinates of point $P$ are $\left(0, \dfrac{25}{6}\right)$