Đáp án:
$\begin{array}{l}
a)\left( d \right)//y = 2x - 1\\
\Rightarrow \left\{ \begin{array}{l}
a = 2\\
b \ne - 1
\end{array} \right.\\
\Rightarrow \left( d \right):y = 2x + b\\
Do:N\left( {1;4} \right) \in \left( d \right)\\
\Rightarrow 4 = 2.1 + b\\
\Rightarrow b = 2\left( {tmdk} \right)\\
Vậy\,y = 2x + 2\\
b)\left( d \right) \bot y = 3x + 2\\
\Rightarrow a.3 = - 1\\
\Rightarrow a = - \dfrac{1}{3}\\
\Rightarrow y = - \dfrac{1}{3}.x + b\\
b = - 2\\
\Rightarrow \left( d \right):y = - \dfrac{1}{3}x - 2\\
c)a = 4\\
\Rightarrow \left( d \right):y = 4x + b\\
M\left( {1;2} \right) \in \left( d \right)\\
\Rightarrow 2 = 4.1 + b\\
\Rightarrow b = - 2\\
Vậy\,\left( d \right)y = 4x - 2\\
d)\left( d \right)//y = x + 1\\
\Rightarrow \left( d \right):y = x + b\\
\left( {0; - 2} \right) \in \left( d \right)\\
\Rightarrow - 2 = 0 + b\\
\Rightarrow b = - 2\\
Vậy\,\left( d \right):y = x - 2\\
e)\left( d \right) \bot y = 3x - 2\\
\Rightarrow a.3 = - 1\\
\Rightarrow a = - \dfrac{1}{3}\\
\Rightarrow \left( d \right):y = - \dfrac{1}{3}x + b\\
y = x - 2\\
+ Khi:x = 1\\
\Rightarrow x - 2 = 1\\
\Rightarrow x = 3\\
\Rightarrow \left( {3;1} \right) \in \left( d \right)\\
\Rightarrow 1 = - \dfrac{1}{3}.3 + b\\
\Rightarrow b = 2\\
Vậy\,\left( d \right):y = - \dfrac{1}{3}x + 2
\end{array}$
