$\mathrm{ABCD}$ is a square . $\mathrm{A}\equiv \left(1,2\right),\mathrm{B}\left(3,-4\right).$ If line CD passes through $(3,8)$, then the midpoint of CD is 

${\mathrm{m}}_{\mathrm{AB}}=\frac{-4-2}{3-1}=-3$

Thus, the equation of CD is $\mathrm{y}-8$ 

$=3\left(\mathrm{x}-3\right),\mathrm{i}.\mathrm{e}.,\mathrm{y}+3\mathrm{x}=17.$

The equation of the right bisector of AB is 

$\mathrm{y}+1=\frac{1}{3}\left(\mathrm{x}-2\right)$

or $3\mathrm{y}=\mathrm{x}-5$

Solving it with line CD, we get $\mathrm{x}=28/5,\mathrm{y}=1/5.$ Thus the midpoint of CD is $\left(28/5,1/5\right)$