Step 1:

The midpoint of the line joining the vertices $(5,7),(-3,-1)$ is the centre of the hyperbola.

(i.e) $c\big(\large\frac{5-3}{2},\large\frac{7-1}{2}\big)$$=c(1,3)$

In the standard form (i.e asymptotes parallel to the coordinate axis),its equation is $(x-1)(y-3)=c^2$

Step 2:

Since the vertices lie on the hyperbola,$(5,7)$ is a point on the rectangular hyperbola.

$(5-1)(7-3)=c^2$

$c^2=16$

Therefore the equation of the rectangular hyperbola is $(x-1)(y-3)=16$

Step 3:

The combined equation of the asymptotes differs only in the constant.

Therefore $(x-1)(y-3)=k$

But the asymptotes pass through the centre $(1,3)$.

Therefore the combined equation of the asymptotes is $(x-1)(y-3)=0$ and the asymptotes are $x=1,y=3$