vecto AB=(1;1)
vecto AM=(x-3;y-1)
\(\left(\overrightarrow{AB},\overrightarrow{AM}\right)=135^0\)
\(\Leftrightarrow\dfrac{1\cdot\left(x-3\right)+1\left(y-1\right)}{\sqrt{1^2+1^2}\cdot\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}}=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\dfrac{x+y-4}{\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}}=-1\)
=>\(x+y-4=-\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}\)
=>\(x=-y+4-\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}\)
AM=2 nên \(\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}=2\)
=>x=-y+4+2=-y+2
=>M(-y+2;y)