Gọi $C(a;b) \\a)\overrightarrow{AC}(a-2;b-1)\\ \overrightarrow{BC}=(a-2;b-3)\\ AC=\sqrt{(a-2)^2+(b-1)^2}\\ BC=\sqrt{(a-2)^2+(b-3)^2}\\ AB=\sqrt{(2-2)^2+(1-3)^2}=4\\ \Delta ABC$ vuông cân tại $C<=>\left\{\begin{array}{l} AC=BC \\ \overrightarrow{AC}.\overrightarrow{BC}=0 \end{array} \right.\\ <=>\left\{\begin{array}{l} \sqrt{(a-2)^2+(b-1)^2}=\sqrt{(a-2)^2+(b-3)^2} \\ (a-2)^2-(b-1)(b-3)=0 \end{array} \right.\\ <=>\left\{\begin{array}{l} (b-1)^2=(b-3)^2 \\ (a-2)^2-(b-1)(b-3)=0 \end{array} \right.\\ <=>\left\{\begin{array}{l} b=2 \\ \left[\begin{array}{l} a=1 \\ a=3 \end{array} \right.\\ \end{array} \right.\\ =>C(1;2); C(3;2)\\ b) \Delta ABC$ đều $=>AC=BC=AB=4\\ <=>\left\{\begin{array}{l} AC=BC \\ AC=4 \end{array} \right.\\ <=>\left\{\begin{array}{l} b=2 \\ \sqrt{(a-2)^2+(b-1)^2}=4 \end{array} \right.\\ <=>\left\{\begin{array}{l} b=2 \\ a=2 \pm \sqrt{15} \end{array} \right.\\ =>C(2 \pm \sqrt{15};2)$
