Find the distance between the parallel lines $5x - 3y + 4 = 0$ and $10x - 6y - 9 = 0$
Equations of the given parallel lines are
$5x - 3y + 4 = 0$ $\;\;\; \cdots \; (1)$; $\;$ $10x - 6y - 9 = 0$ $\;\;\; \cdots \; (2)$
The perpendicular distance from $\left(0,0\right)$ to equation $(1)$ is
$= \left|\dfrac{4}{\sqrt{5^2 + \left(-3\right)^2}}\right| = \dfrac{4}{\sqrt{34}}$
The perpendicular distance from $\left(0,0\right)$ to equation $(2)$ is
$= \left|\dfrac{-9}{\sqrt{10^2 + \left(-6\right)^2}}\right| = \dfrac{9}{\sqrt{136}} = \dfrac{9}{2 \sqrt{34}}$
Substituting $y =0$ in equation $(1)$ gives $\;\;$ $x = \dfrac{-4}{5}$
$\implies$ Equation $(1)$ cuts off negative intercept on the $X$ axis.
Substituting $y =0$ in equation $(2)$ gives $\;\;$ $x = \dfrac{9}{10}$
$\implies$ Equation $(2)$ cuts off positive intercept on the $X$ axis.
$\therefore \;$ The origin lies between the lines given by equations $(1)$ and $(2)$.
$\therefore \;$ The distance between the parallel lines is
$= \dfrac{4}{\sqrt{34}} + \dfrac{9}{2 \sqrt{34}} = \dfrac{17}{2 \sqrt{34}} = \dfrac{34}{4 \sqrt{34}} = \dfrac{\sqrt{34}}{4}$