If ${\mathrm{Z}}_1=2-3\mathrm{i}$ and ${\mathrm{Z}}_2=-1+\mathrm{i}$  then the locus of
a point  represented by Z = x + iy in the argand
plane satisfying the equation

$\arg \left(\frac{\mathrm{Z}-{\mathrm{Z}}_1}{\mathrm{Z}-{\mathrm{Z}}_2}\right)=\frac{\mathrm{\pi }}{2}$

$$\begin{gathered} {\mathrm{z}}_1=2-3\mathrm{i}, {\mathrm{z}}_2=-1+\mathrm{i} \mathrm{and} \\ \mathrm{Arg} \left(\frac{\mathrm{z}-{\mathrm{z}}_1}{\mathrm{z}-{\mathrm{z}}_2}\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \mathrm{Arg} \left(\frac{\mathrm{z}-(2-3\mathrm{i})}{\mathrm{z}-(-1+\mathrm{i})}\right)=\mathrm{Arg} \left(\frac{\mathrm{x}+\mathrm{iy}-(2-3\mathrm{i})}{\mathrm{x}+\mathrm{iy}-(-1+\mathrm{i})}\right)=\frac{\mathrm{\pi }}{2} \\ \therefore \mathrm{Arg} \left(\frac{(\mathrm{x}-2)+\mathrm{i}(\mathrm{y}+3)}{(\mathrm{x}+1)+\mathrm{i}(\mathrm{y}-1)}\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\mathrm{Tan}}^{-1} \frac{\mathrm{y}+3}{\mathrm{x}-2}-{\mathrm{Tan}}^{-1} \frac{\mathrm{y}-1}{\mathrm{x}+1}=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \frac{{\displaystyle \frac{\mathrm{y}+3}{\mathrm{x}-2}}-{\displaystyle \frac{\mathrm{y}-1}{\mathrm{x}+1}}}{1+\left({\displaystyle \frac{\mathrm{y}+3}{\mathrm{x}-2}}\right) \left({\displaystyle \frac{\mathrm{y}-1}{\mathrm{x}+1}}\right)}=\infty \\ \Rightarrow \frac{(\mathrm{x}+1)(\mathrm{y}+3)-(\mathrm{y}-1)(\mathrm{x}-2)}{(\mathrm{x}+1)(\mathrm{x}-2)+(\mathrm{y}+3)(\mathrm{y}-1)}=\infty \\ \Rightarrow \frac{4\mathrm{x}+3\mathrm{y}+1}{{\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}+2\mathrm{y}-5}=\infty \\ \Rightarrow \boxed{{\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}+2\mathrm{y}-5=0, 4\mathrm{x}+3\mathrm{y}+1>0} \end{gathered}$$