Reduce the equation $\;$ $3x + 4y = 5$ $\;$ to the slope-intercept form, the intercept form and the perpendicular form.
Equation of given straight line: $\;$ $3x + 4y = 5$ $\;\;\; \cdots \; (1)$
Slope-intercept form:
Equation $(1)$ can be written as
$4y = -3x + 5$
i.e. $\;$ $y = \dfrac{-3}{4} x + \dfrac{5}{4}$ $\;\;\; \cdots \; (2)$
Equation $(2)$ is the slope-intercept form of equation $(1)$ with slope $= m = \dfrac{-3}{4}$ and intercept $= c = \dfrac{5}{4}$
Intercept form:
Rearranging equation $(1)$ we have,
$\dfrac{3x}{5} + \dfrac{4 y}{5} = 1$
i.e. $\;$ $\dfrac{x}{5 / 3} + \dfrac{y}{5 / 4} = 1$ $\;\;\; \cdots \; (3)$
Equation $(3)$ is the intercept form of equation $(1)$ with intercept on the $X$ axis $= a = \dfrac{5}{3}$ and intercept on the $Y$ axis $= b = \dfrac{5}{4}$
Perpendicular form:
Writing equation $(1)$ as
$3x + 4y - 5 = 0$
i.e. $\;$ $\dfrac{3x}{\sqrt{3^2 + 4^2}} + \dfrac{4y}{\sqrt{3^2 + 4^2}} - \dfrac{5}{\sqrt{3^2 + 4^2}} = 0$
i.e. $\;$ $\dfrac{3x}{5} + \dfrac{4y}{5} - \dfrac{5}{5} = 0$
i.e. $\;$ $x \cos \alpha + y \sin \alpha -1 = 0$ $\;\;\; \cdots \; (4)$
where $\;$ $\cos \alpha = \dfrac{3}{5}$, $\;$ $\sin \alpha = \dfrac{4}{5}$ $\;$ i.e. $\;$ $\tan \alpha = \dfrac{4}{3}$
Equation $(4)$ is the required perpendicular form of equation $(1)$.