Since $a = bm, $ we can substitute $bm$ into the variable $a$ in the second equation: $$(bm)^2 + b^2 = c^2\implies b^2(m^2 + 1) = c^2 \implies b^2 = \dfrac{c^2}{m^2 + 1} \implies b = \dfrac{c}{\sqrt{m^2 + 1}}.$$ Since the lengths of the sides of a triangle must be positive, we can take the positive root of $b^2$ to solve for $b$, then back substitute to obtain $a = bm$.
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If you have the "slope"
$$m = \frac ab$$
then you can write $a$ as $mb$. Fit this in
$$c = \sqrt{a^2 + b^2}$$
and get
$$b = \frac{c}{\sqrt{m^2 + 1}}$$
We have a right triangle, so there are two things we know:
And
Two equations and two unknowns.
SPOILER ALERT: