Ta có
a) \(M\in Ox\Rightarrow M\left(x;0\right)\)
\(\overrightarrow{AM}=\left(x+1;2\right)\Rightarrow AM=\sqrt{x^2+2x+5}\Rightarrow AM^2=x^2+2x+5\)
\(\overrightarrow{BM}=\left(x-4;-1\right)\Rightarrow BM=\sqrt{x^2-8x+17}\Rightarrow BM^2=x^2-8x+17\)
Vì AM = BM nên \(AM^2=BM^2\)
\(\Leftrightarrow x^2+2x+5=x^2-8x+17\Leftrightarrow10x=12\Leftrightarrow x=\frac{6}{5}\)
Vậy \(M\left(\frac{6}{5};0\right)\)
b) \(N\in Oy\Rightarrow N\left(0;y\right)\)
\(\overrightarrow{NA}=\left(-1;-2-y\right)\)
\(\overrightarrow{NB}=\left(4;1-y\right)\)
Vì \(\Delta ABN\) vuông tại N
\(\Rightarrow\overrightarrow{NA}.\overrightarrow{NB}=0\Leftrightarrow\left(-1\right)\times4+\left(-2-y\right)\left(1-y\right)=0\Leftrightarrow y^2+y-6=0\Leftrightarrow y=3;y=-2\)
Vậy \(N\left(0;3\right)\) hoặc \(N\left(0;-2\right)\)
