Đáp án:
Gọi pt đường thẳng cần tìm có dạng: y=a.x+b
$\begin{array}{l}
a)\left( d \right):x - y + 1 = 0\\
\Rightarrow \left( d \right):y = x + 1\\
Do:\left( \Delta \right)//\left( d \right)\\
\Rightarrow \left\{ \begin{array}{l}
a = 1\\
b \ne 1
\end{array} \right.\\
\Rightarrow \left( \Delta \right):y = x + b\\
Do:A\left( {1;3} \right) \in \left( \Delta \right)\\
\Rightarrow 3 = 1 + b\\
\Rightarrow b = 2\left( {tmdk} \right)\\
Vậy\,\left( \Delta \right):y = x + 2\\
b)\left( d \right):2x + y - 1 = 0\\
\Rightarrow y = - 2x + 1\\
\left( \Delta \right)//\left( d \right)\\
\Rightarrow \left\{ \begin{array}{l}
a = - 2\\
b \ne 1
\end{array} \right.\\
\Rightarrow \left( \Delta \right):y = - 2x + b\\
A\left( { - 1;0} \right) \in \left( \Delta \right)\\
\Rightarrow 0 = - 2.\left( { - 1} \right) + b\\
\Rightarrow b = - 3\left( {tmdk} \right)\\
Vậy\,\left( \Delta \right):y = - 2x - 3\\
c)A\left( {3;2} \right);\left( \Delta \right)//Ox\\
Ox:y = 0\\
\Rightarrow \left( \Delta \right):y = a\left( {a \ne 0} \right)\\
Do:A\left( {3;2} \right) \in \left( \Delta \right)\\
\Rightarrow \left( \Delta \right)y = 2\\
Vậy\,\left( \Delta \right):y = 2
\end{array}$
