#### Answer

$$(b-a) \sqrt{1+m^{2}}$$

#### Work Step by Step

By the Pythagorean Theorem
\begin{array}{l}
{s^{2}=(b-a)^{2}+(m(b-b))^{2}} \\
{\left.s=(c b-a)^{2}+m^{2}(b-a)^{2}\right)^{\frac{1}{2}}} \\
{=(b-a)\left(1 - m^{2}\right)^{\frac{1}{2}}}
\end{array}
Then, by using arc length
\begin{aligned}
s &=\int_{a}^{b} \sqrt{1+m^{2}} d x \\
&=[x \sqrt{1+m^{2}}]_{a}^{b} \\
&=b \sqrt{1+m^{2}}-a \sqrt{1+m^{2}} \\
&=(b-a) \sqrt{1+m^{2}}
\end{aligned}