Cho hai đường thẳng \(\left( {{d_1}} \right):mx + y = m + 1\,\,,\left( {{d_2}} \right):x + my = 2\,\)song song nhau khi và chỉ khi

+) Nếu \(m = 0\) thì \({d_1}:y = 1,{d_2}:x = 2\) cắt nhau tại \(\left( {2;1} \right)\).

+) Nếu \(m \ne 0\) thì \({d_1}//{d_2} \Leftrightarrow \dfrac{m}{1} = \dfrac{1}{m} \ne \dfrac{{m + 1}}{2} \)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}\dfrac{m}{1} = \dfrac{1}{m}\\\dfrac{1}{m} \ne \dfrac{{m + 1}}{2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}{m^2} = 1\\1.2 \ne m\left( {m + 1} \right)\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m =  \pm 1\\{m^2} + m - 2 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m =  \pm 1\\m \ne 1\\m \ne 2\end{array} \right. \\\Leftrightarrow m =  - 1\end{array}\)