Gọi \(M\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AM}=\left(x+1;y+3\right)\\\overrightarrow{AB}=\left(1;-1\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}AM=\sqrt{\left(x+1\right)^2+\left(y+3\right)^2}\\AB=\sqrt{2}\end{matrix}\right.\)
Tam giác ABM vuông tại A và có diện tích 4
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AM}.\overrightarrow{AB}=0\\\dfrac{1}{2}AM.AB=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+1-\left(y+3\right)=0\\\sqrt{2\left(x+1\right)^2+2\left(y+3\right)^2}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\\left(x+1\right)^2+\left(y+3\right)^2=32\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2+\left(x-2+3\right)^2=32\)
\(\Leftrightarrow\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x=3\Rightarrow y=1\\x=-5\Rightarrow y=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}M\left(3;1\right)\\M\left(-5;-7\right)\end{matrix}\right.\)
