\(\overrightarrow{AB}=\left(-3;4\right)\Rightarrow\) đường thẳng AB nhận \(\left(4;3\right)\) là 1 vtpt
Phương trình AB:
\(4\left(x-3\right)+3\left(y+1\right)=0\Leftrightarrow4x+3y-9=0\)
Gọi \(M\left(m;0\right)\Rightarrow d\left(M;AB\right)=1\)
\(\Leftrightarrow\frac{\left|4m-9\right|}{\sqrt{4^2+3^2}}=1\Leftrightarrow\left|4m-9\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}4m-9=5\\4m-9=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}M\left(\frac{7}{2};0\right)\\M\left(1;0\right)\end{matrix}\right.\)