ABC is an isosceles triangle and $\mathrm{B} = {90}^0$. If B and the midpoint P of AC are represented by $3 + 2\mathrm{i} \mathrm{and} 1 - \mathrm{i}$ then the other vertices are

$$\begin{gathered} \mathrm{B}(3, 2) \mathrm{midpoint} \mathrm{of} \mathrm{AC} = M(1, -1), \\ \mathrm{AB} = \mathrm{AC} \Rightarrow BM \perp \mathrm{AC}, \\ \mathrm{Slope} \mathrm{of} \mathrm{BM} = \frac{2+ 1}{3-1} = \frac{3}{2} \\ \Rightarrow \mathrm{slope} \mathrm{of} \mathrm{AC} = \frac{-2}{3}, \tan \mathrm{\theta } = \frac{-2}{3}. \\ \Rightarrow \cos \mathrm{\theta } = \frac{-3}{\sqrt{13}}, \sin \mathrm{\theta } = \frac{2}{\sqrt{13}}, \\ \mathrm{M} = \mathrm{Circum} \mathrm{centre} \Rightarrow \mathrm{MB} = \mathrm{MA} = \mathrm{MC} \\ and \mathrm{MB} = \sqrt{{(3-1)}^2 + {(2 + 1)}^2} = \sqrt{13} \\ \mathrm{Vertices} = ({\mathrm{x}}_1 \pm r\cos \mathrm{\theta }, {\mathrm{y}}_1 \pm r\mathrm{sin\theta } ) = (1 \pm \sqrt{13}(\frac{-3}{\sqrt{13}}) , -1 \pm \sqrt{13}\left(\frac{2}{\sqrt{13}}\right)) \\ = (-2, 1) \mathrm{and} (4 , -3) \end{gathered}$$