$OACB$ is a rectangle where $OA = a$, $OB = b$. Taking $OA$, $OB$ as axes of coordinates, prove that the equation of $AE$ where $E$ is the midpoint of $BC$ is $\;$ $\dfrac{x}{a} + \dfrac{y}{2b} = 1$.
Given: $\;$ $OA$ and $OB$ are the axes of coordinates.
Let $OA$ represent the $X$ axis and $OB$ represent the $Y$ axis.
Let $O$ be the origin $\left(0,0\right)$.
Given: $\;$ $OA = a$ and $OB = b$ $\implies$ $A = \left(a, 0\right)$ and $B = \left(0, b\right)$
Then $\;$ $C = \left(a, b\right)$
$E$ is the midpoint of $BC$.
$\therefore \;$ The coordinates of $E = \left(\dfrac{a + 0}{2}, \dfrac{b + b}{2}\right) = \left(\dfrac{a}{2}, b\right)$
Slope of $AE = \dfrac{b - 0}{\dfrac{a}{2} - a} = \dfrac{-2b}{a}$
$\therefore \;$ Equation of $AE$ is
$y - 0 = \dfrac{-2b}{a} \left(x - a\right)$
i.e. $\;$ $ay = -2bx + 2ab$
i.e. $\;$ $\dfrac{y}{2b} = - \dfrac{x}{a} + 1$
i.e. $\;$ $\dfrac{x}{a} + \dfrac{y}{2b} = 1$