Find the perpendicular distance from the point $\left(3, 4\right)$ to the straight line $3x - 4y + 10 = 0$
Equation of given line is $\;$ $3x - 4y + 10 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the standard equation of a straight line $Ax + By + C = 0$ gives
$A = 3$, $\;$ $B = -4$, $\;$ $C = 10$
Given point: $\left(x_1, y_1\right) = \left(3, 4\right)$
Perpendicular distance from the point $\left(x_1, y_1\right)$ to the line $Ax + By + C = 0$ is $= \dfrac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}$
$\therefore \;$ Required perpendicular distance is
$= \dfrac{3 \times 3 - 4 \times 4 + 10}{\sqrt{3^2 + \left(-4\right)^2}} = \dfrac{9 - 16 + 10}{\sqrt{25}} = \dfrac{3}{5}$