Find the perpendicular distance between the parallel lines $3x-4y=12$ and $3x-4y=24$.

Step-by-step solution

1. Both lines have the same coefficients $a=3$, $b=-4$; only constants differ.
2. Compute the difference of constants: $|24-12|=12$.
3. Compute $\sqrt{a^2+b^2}=\sqrt{3^2+(-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5$.
4. Apply distance formula for parallel lines: $d=\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}=\frac{12}{5}=2.4$ units.
5. Thus the distance is $\frac{12}{5}$ units.

Concept tested: distance between parallel lines

Published 13 October 2026 · How this solution was prepared and checked