if A(9,-9), (1,-3) are the vertices of right angled isosceles triangle, then find the third vertex

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$$\begin{gathered} \mathrm{I} \mathrm{am} \mathrm{assuming} \mathrm{that} \mathrm{given} \mathrm{vertices} \mathrm{are} \mathrm{vertices} \mathrm{of} \mathrm{hypotenuse}. \\ \mathrm{Let} \mathrm{A}\left(9,-9\right) \mathrm{and} \mathrm{B}\left(1,-3\right) \\ \mathrm{Let} \mathrm{third} \mathrm{vertex} \mathrm{be} \mathrm{C}\left(\mathrm{h},\mathrm{k}\right) \\ \mathrm{As} \mathrm{triangle} \mathrm{is} \mathrm{isosceles}, \mathrm{hence} \\ \mathrm{AC}=\mathrm{BC} \\ \Rightarrow \sqrt{{\left(\mathrm{h}-9\right)}^2+{\left(\mathrm{k}+9\right)}^2}=\sqrt{{\left(\mathrm{h}-1\right)}^2+{\left(\mathrm{k}+3\right)}^2} \\ \Rightarrow {\left(\mathrm{h}-9\right)}^2+{\left(\mathrm{k}+9\right)}^2={\left(\mathrm{h}-1\right)}^2+{\left(\mathrm{k}+3\right)}^2 \\ \Rightarrow {\mathrm{h}}^2+81-18\mathrm{h}+{\mathrm{k}}^2+81+18\mathrm{h}={\mathrm{h}}^2+1-2\mathrm{h}+{\mathrm{k}}^2+9+6\mathrm{k} \\ \Rightarrow 81-18\mathrm{h}+81+18\mathrm{k}=1-2\mathrm{h}+9+6\mathrm{k} \\ \Rightarrow 162-10+18\mathrm{k}-6\mathrm{k}=18\mathrm{h}-2\mathrm{h} \\ \Rightarrow 152+12\mathrm{k}=16\mathrm{h} \\ \Rightarrow 38+3\mathrm{k}=4\mathrm{h} \\ \Rightarrow \mathrm{h}=\frac{3\mathrm{k}+38}{4} ;\left(\mathrm{i}\right) \\ \mathrm{As} \mathrm{triangle} \mathrm{is} \mathrm{right} \mathrm{angled}, \mathrm{hence} \\ \mathrm{AC} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{BC} \\ \mathrm{Slope} \mathrm{of} \mathrm{AC}\times \mathrm{Slope} \mathrm{of} \mathrm{BC}=-1 \\ \Rightarrow \left(\frac{\mathrm{k}+9}{\mathrm{h}-9}\right)\left(\frac{\mathrm{k}+3}{\mathrm{h}-1}\right)=-1 \\ \mathrm{Now} \mathrm{we} \mathrm{will} \mathrm{put} \mathrm{value} \mathrm{of} \mathrm{h} \mathrm{from} \left(\mathrm{i}\right) \mathrm{to} \mathrm{obtain} \mathrm{value} \mathrm{of} \mathrm{k} \end{gathered}$$

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