Find the equation of the straight line parallel to $5x - 4y + 3 = 0$ and having X intercept 3.
Equation of the given line is: $\hspace{1em}$ $5x - 4 y + 3 = 0$
i.e. $y = \dfrac{5}{4}x + \dfrac{3}{4}$
$\therefore$ $\;$ Slope of the given line $= m = \dfrac{5}{4}$
Since the required line is parallel to the given line,
slope of required line $= m = \dfrac{5}{4}$
Let the intercept made by the required line on the X axis be $a$.
Given: $a = 3$
Let the required line make an angle $\alpha$ with the X axis.
Then, $\tan \alpha = \dfrac{PB}{PA} = \dfrac{y}{x - a}$
But $\tan \alpha$ gives the slope of the line.
$\therefore$ $\;$ Slope of the required line $ = m = \dfrac{y}{x - a}$
i.e. $y = m \left(x - a\right)$ $\;\;\; \cdots \; (1)$
Substituting the values of $m$ and $a$ in equation $(1)$, we have
$y = \dfrac{5}{4} \left(x - 3\right)$
i.e. $4y = 5x - 15$
or, $5x - 4y - 15 = 0$
which is the equation of the required line.