smc

Notice that since $(a,b)$ is on $y=mx+k$, we have $b=am+k$. Similarly, $d=cm+k$. Using the distance formula, the distance between the points $(a,b)$ and $(c,d)$ is $\sqrt{(a-c{)}^2+(b-d{)}^2}=\sqrt{(a-c{)}^2+(am+k-cm-k{)}^2}$ $=\sqrt{(a-c{)}^2+m^2(a-c{)}^2}$ $=|a-c|\sqrt{1+m^2}.$ And so the answer is $\boxed{|a-c|\sqrt{1+m^2}}$.