Find the equation of the straight line parallel to the lines $2x + 3y = 5$ and $2x + 3y + 4 = 0$, which divides the distance between them internally in the ratio $3 : 5$.
The given lines are: $\;$ $2x + 3y = 5$ $\;\;\; \cdots \; (1)$; $\;$ $2x + 3y + 4 = 0$ $\;\;\; \cdots \; (2)$
Putting $y = 0$ in equation $(1)$ gives $\;\;$ $x = \dfrac{5}{2}$
$\implies$ Equation $(1)$ cuts the $X$ axis at the point $A \left(\dfrac{5}{2}, 0\right)$
Putting $y = 0$ in equation $(2)$ gives $\;\;$ $x = -2$
$\implies$ Equation $(2)$ cuts the $X$ axis at the point $B \left(-2, 0\right)$
Let $P \left(x_1, 0\right)$ be a point on the required line which cuts the line segment $AB$ internally in the ratio $3 : 5$
Then by section formula for internal division,
$x_1 = \dfrac{3 \times \left(-2\right) + 5 \times \dfrac{5}{2}}{3 + 5} = \dfrac{-6 + \dfrac{25}{2}}{8} = \dfrac{13}{16}$
i.e. $\;$ $P \left(x_1, 0\right) = \left(\dfrac{13}{16}\right)$
Slope of line given by equation $(1)$ is $m = \dfrac{-2}{3}$
Since the required line is parallel to lines given by equations $(1)$ and $(2)$,
slope of required line $= m = \dfrac{-2}{3}$
and the required line passes through the point $P$.
$\therefore \;$ The equation of the required line is
$y - 0 = \dfrac{-2}{3} \left(x - \dfrac{13}{16}\right)$
i.e. $\;$ $3y = -2x + \dfrac{13}{8}$
i.e. $\;$ $16 x + 24 y - 13 = 0$