Do MN là đường trung bình \(\Rightarrow MN//AB\Rightarrow\) pt MN có dạng:
\(1\left(x-2\right)-1\left(y-1\right)=0\Leftrightarrow x-y-1=0\)
\(S_{ABC}=\frac{1}{2}AB.d\left(C;AB\right)=MN.d\left(C;AB\right)=2MN.d\left(M;AB\right)\)
Mà \(d\left(M;AB\right)=\frac{\left|2-1\right|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow2MN.\frac{1}{\sqrt{2}}=2\Rightarrow MN=\sqrt{2}\)
Gọi N \(\left(a;a-1\right)\) \(\Rightarrow MN^2=\left(a-2\right)^2+\left(a-2\right)^2=2\)
\(\Rightarrow\left(a-2\right)^2=1\Rightarrow\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}N\left(3;2\right)\\N\left(1;0\right)\end{matrix}\right.\)
