Find the perpendicular distance from the point $(3, 4)$ to the line $3x + 4y - 5 = 0$.

Use $d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} = \dfrac{|3(3) + 4(4) - 5|}{\sqrt{9 + 16}} = \dfrac{20}{5} = 4$. Omitting the denominator leaves 20, and dividing by $a + b = 7$ instead of $\sqrt{a^2 + b^2}$ gives $\dfrac{20}{7}$.