Đáp án:
$\begin{array}{l}
1)a)\left( d \right)//\Delta \\
\Rightarrow \left( d \right): - x - 2y + c = 0\\
A\left( {1;2} \right) \in \left( d \right)\\
\Rightarrow - 1 - 2.2 + c = 0\\
\Rightarrow c = 5\\
\Rightarrow \left( d \right): - x - 2y + 5 = 0\\
hay\,\left( d \right):x + 2y - 5 = 0\\
b)\left( d \right) \bot \Delta \\
\Rightarrow \left( d \right):2x - y + d = 0\\
B\left( { - 2;1} \right) \in \left( d \right)\\
\Rightarrow \left( d \right):2.\left( { - 2} \right) - 1 + d = 0\\
\Rightarrow d = 5\\
\Rightarrow \left( d \right):2x - y + 5 = 0\\
B2)\\
a)d \equiv Ox \Rightarrow d:y = 0\\
b)d//Oy:x = 0\\
\Rightarrow d:x = a\\
M\left( {2; - 1} \right) \in d\\
\Rightarrow x = 2\\
c)d:y = a.x + b\\
Do:A\left( { - 2;0} \right);B\left( {0;3} \right) \in d\\
\Rightarrow \left\{ \begin{array}{l}
0 = a.\left( { - 2} \right) + b\\
3 = a.0 + b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2a = b\\
b = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = \dfrac{3}{2}\\
b = 3
\end{array} \right.\\
Vậy\,\left( d \right):y = \dfrac{3}{2}.x + 3
\end{array}$
