\(\overrightarrow{AB}=\left(4;2\right)=2\left(2;1\right)\Rightarrow AB=2\sqrt{5}\)
Đường thẳng AB nhận (1;-2) là 1 vtpt nên pt có dạng:
\(1\left(x+1\right)-2\left(y-2\right)=0\Leftrightarrow x-2y+5=0\)
\(\overrightarrow{AC}=\left(2;-8\right)=2\left(1;-4\right)\Rightarrow AC=2\sqrt{17}\)
Đường thẳng AC nhận (4;1) là 1 vtpt nên pt có dạng:
\(4\left(x+1\right)+1\left(y-2\right)=0\Leftrightarrow4x+y+2=0\)
Gọi \(M\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}S_{MAB}=\dfrac{1}{2}d\left(M;AB\right).AB\\S_{MAC}=\dfrac{1}{2}d\left(M;AC\right).AC\end{matrix}\right.\)
\(S_{MAB}=S_{MAC}=d\left(M;AB\right).AB=d\left(M;AC\right).AC\)
\(\Leftrightarrow\dfrac{\left|x-2y+5\right|}{\sqrt{1+\left(-2\right)^2}}.2\sqrt{5}=\dfrac{\left|4x+y+2\right|}{\sqrt{4^2+1^2}}.2\sqrt{17}\)
\(\Leftrightarrow\left|x-2y+5\right|=\left|4x+y+2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+y+2=x-2y+5\\4x+y+2=-x+2y-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y-1=0\\5x-y+7=0\end{matrix}\right.\)
Vậy quỹ tích M là 2 đường thẳng có pt: \(\left[{}\begin{matrix}x+y-1=0\\5x-y+7=0\end{matrix}\right.\)