The middle point of line segment joining (3, -1) and (1, 1) is shifted by 2 units (increasing y) perpendicular to line segment. Coordinates of point in new position is

Let A (3, -1 )  B(1, 1)  and  r=2

               P = mid point AB $=\left(\frac{3+1}{2},\frac{-1+1}{2}\right)=\left(2,0\right)=\left(x_1,y_1\right)$

Slope of AB $=\frac{1+1}{1-3}=\frac{2}{-2}=-1$

$\begin{array}{l}Slopeof\perp rline\overline{PQ}=1\\ Let\theta betheanglemadeby\overline{PQ}withx-axis\\ \therefore \tan \theta =1\Rightarrow \theta ={45}^0\\ \therefore Q=\left(x_1+r\cos \theta ,y_1+r\sin \theta \right)\\ \Rightarrow \left(2+2\cos \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.,0+2\sin \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)\\ \Rightarrow \left(2+2\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.\right),0+2\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.\right)\right)\\ \Rightarrow \left(2+\sqrt{2},\sqrt{2}\right)\end{array}$