Đáp án:
$\begin{array}{l}
a)A\left( { - 1;2} \right);B\left( {3; - 4} \right) \in \left( d \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
2 = \left( {m - 2} \right).\left( { - 1} \right) + n\\
- 4 = \left( {m - 2} \right).3 + n
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- m + 2 + n = 2\\
3m - 6 + n = - 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m - n = 0\\
3m + n = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = \dfrac{1}{2}\\
n = \dfrac{1}{2}
\end{array} \right.\\
Vậy\,m = n = \dfrac{1}{2}\\
b)\left( d \right) \cap Oy = \left( {0;1 - \sqrt 2 } \right)\\
\left( d \right) \cap Ox = \left( {2 + \sqrt 2 ;0} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
1 - \sqrt 2 = \left( {m - 2} \right).0 + n\\
0 = \left( {m - 2} \right).\left( {2 + \sqrt 2 } \right) + n
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
n = 1 - \sqrt 2 \\
\left( {m - 2} \right).\left( {2 + \sqrt 2 } \right) + 1 - \sqrt 2 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
n = 1 - \sqrt 2 \\
m - 2 = \dfrac{{\sqrt 2 - 1}}{{2 + \sqrt 2 }} = \dfrac{{ - 4 + 3\sqrt 2 }}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
n = 1 - \sqrt 2 \\
m = \dfrac{{3\sqrt 2 }}{2}
\end{array} \right.\\
Vậy\,m = \dfrac{{3\sqrt 2 }}{2};n = 1 - \sqrt 2 \\
c)\left( d \right)// - 2y + x - 3 = 0\\
Do: - 2y + x - 3 = 0\\
\Leftrightarrow 2y = x - 3\\
\Leftrightarrow y = \dfrac{1}{2}.x - \dfrac{3}{2}\\
\Leftrightarrow m = \dfrac{1}{2}\\
Vậy\,m = \dfrac{1}{2}\\
d)3x + 2y = 1\\
\Leftrightarrow 2y = - 3x + 1\\
\Leftrightarrow y = - \dfrac{3}{2}.x + \dfrac{1}{2}\\
Khi:\left( d \right) \cap y = - \dfrac{3}{2}.x + \dfrac{1}{2}\\
\Leftrightarrow m\# - \dfrac{3}{2}\\
Vậy\,m\# - \dfrac{3}{2}
\end{array}$
