Đáp án:
Gọi pt đường thẳng d có dạng: y=a.x+b
$\begin{array}{l}
a)P\left( {3; - 2} \right),N\left( {1;2} \right) \in d\\
\Leftrightarrow \left\{ \begin{array}{l}
- 2 = 3a + b\\
2 = a + b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2a = - 4\\
b = 2 - a
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = - 2\\
b = 4
\end{array} \right.\\
Vậy\,\left( d \right):y = - 2x + 4\\
b)d//y = 2x - 1\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 2\\
b \ne - 1
\end{array} \right.\\
\Leftrightarrow d:y = 2x + b\\
F\left( { - 1;1} \right) \in d\\
\Leftrightarrow 1 = 2.\left( { - 1} \right) + b\\
\Leftrightarrow b = 3\left( {tmdk} \right)\\
Vậy\,\left( d \right)y = 2x + 3\\
c)a = - 3\\
\Leftrightarrow d:y = - 3x + b\\
E\left( { - 2;3} \right) \in d\\
\Leftrightarrow 3 = - 3.\left( { - 2} \right) + b\\
\Leftrightarrow b = - 3\\
Vậy\,\left( d \right):y = - 3x - 3\\
d)d \bot \Delta :y = 2x + 1\\
\Leftrightarrow a.2 = - 1\\
\Leftrightarrow a = - \dfrac{1}{2}\\
\Leftrightarrow d:y = - \dfrac{1}{2}x + b\\
Q\left( { - 1; - 3} \right) \in d\\
\Leftrightarrow - 3 = - \dfrac{1}{2}.\left( { - 1} \right) + b\\
\Leftrightarrow b = - 3 - \dfrac{1}{2} = - \dfrac{7}{2}\\
Vậy\,\left( d \right):y = - \dfrac{1}{2}x - \dfrac{7}{2}
\end{array}$
