Before moving on to solve this question, I need to make you understand that “the locus of point which is equidistant from two fix point two is the perpendicular bisector of the fixed two points.”   

Let the bisector of PQ = R

Mid-point of  PQ is $R=\left( \frac{1+3}{2},\frac{3+5}{2} \right)=(2,4)$

${{m}_{1}}=slope\text{ of }PQ=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{5-3}{3-1}=\frac{2}{2}=1$

Perpendicular bisector to PQ is RX

For two lines to be perpendicular ${{m}_{1}}{{m}_{2}}=-1$

${{m}_{2}}=\frac{-1}{1}=-1$

Using point R (2,4)

Using one point slope form equation of a line

$(y-{{y}_{1}})={{m}_{2}}(x-{{x}_{1}})$

$\text{Take }R\text{ as }(2,4)$

$y-4=-1(x-2)$

$y-4=-x+2$

$y=-x-6$