Gọi \(C\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AC}=\left(x-2;y-1\right)\\\overrightarrow{AB}=\left(-3;-1\right)\end{matrix}\right.\)
Để ABC vuông cân tại A
\(\Leftrightarrow\left\{{}\begin{matrix}AB^2=AC^2\\\overrightarrow{AB}.\overrightarrow{AC}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2+\left(y-1\right)^2=10\\3\left(x-2\right)=y-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\\left(x-2\right)^2+\left(3x-6\right)^2=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-5\\\left(x-2\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1;y=-2\\x=3;y=1\end{matrix}\right.\)
Có 2 điểm C thỏa mãn: \(\left[{}\begin{matrix}C\left(1;-2\right)\\C\left(3;1\right)\end{matrix}\right.\)