Đáp án: $\left[\begin{array}{l}M\left(3;4\right)\\M\left(2;1\right)\end{array}\right.$
Giải thích:
Gọi $M\left( x;y \right)$
$\begin{cases}\overrightarrow{AM}\left( x-1;y-3 \right)\to A{{M}^{2}}={{\left( x-1 \right)}^{2}}+{{\left( y-3 \right)}^{2}}\\\overrightarrow{BM}\left( x-4;y-2 \right)\to B{{M}^{2}}={{\left( x-4 \right)}^{2}}+{{\left( y-2 \right)}^{2}}\end{cases}$
Để $\Delta ABM$ vuông cân tại $M$ thì:
$\begin{cases}AM=BM\\\overrightarrow{AM}.\overrightarrow{BM}=0\end{cases}$
$\bullet \,\,\,\,\,AM=BM$
$\Leftrightarrow A{{M}^{2}}=B{{M}^{2}}$
$\Leftrightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-3 \right)}^{2}}={{\left( x-4 \right)}^{2}}+{{\left( y-2 \right)}^{2}}$
$\Leftrightarrow {{x}^{2}}-2x+1+{{y}^{2}}-6y+9={{x}^{2}}-8x+16+{{y}^{2}}-4y+4$
$\Leftrightarrow 6x-2y=10\,$
$\Leftrightarrow 3x-y=5$
$\Leftrightarrow y=3x-5$
$\bullet \,\,\,\,\,\overrightarrow{AM}.\overrightarrow{BM}=0$
$\Leftrightarrow \left( x-1 \right)\left( x-4 \right)+\left( y-2 \right)\left( y-3 \right)=0$
$\Leftrightarrow \left( x-1 \right)\left( x-4 \right)+\left( 3x-7 \right)\left( 3x-8 \right)=0$
$\Leftrightarrow {{x}^{2}}-5x+4+9{{x}^{2}}-45x+56=0$
$\Leftrightarrow 10{{x}^{2}}-50x+60=0$
$\Leftrightarrow\left[\begin{array}{l}x=3\\x=2\end{array}\right.$
Với $x=3\to y=4$
Với $x=2\to y=1$
Vậy $M\left( 3;4 \right)$ hoặc $M\left( 2;1 \right)$
