Perpendicular Line

The midpoint of segment $AB$ can be computed as:

$P(\bar{x},\bar{y}) = P(\frac{1+5}{2},\frac{a+b}{2}) = P(3, \frac{a+b}{2}).$

If the perpendicular bisector of $AB$ is the line $y = \frac{1}{3}x$ , then we have for the slope of $AB$ :

$\frac{b-a}{5-1} = -3 \Rightarrow b-a = -12$ (i)

and if the perpendicular bisector includes the midpoint $P$ above, then we have:

$\frac{a+b}{2} = \frac{1}{3}(3) \Rightarrow a+b=2$ (ii).

Solving for $a$ and $b$ in (i) and (ii) above gives us $a = 7, b = -5 \Rightarrow a \cdot b = \boxed{-35}.$