Đáp án:
$\begin{array}{l}
a)y = a.x + b\\
A\left( { - 3; - 4} \right);B\left( {0;4} \right) \in y = a.x + b\\
\Leftrightarrow \left\{ \begin{array}{l}
- 4 = a.\left( { - 3} \right) + b\\
4 = a.0 + b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 3a + b = - 4\\
b = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 3a = - 4 - b = - 8\\
b = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = \dfrac{8}{3}\\
b = 4
\end{array} \right.\\
Vậy\,a = \dfrac{8}{3};b = 4\\
c)y = a.x + b//y = - \dfrac{2}{3}x + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
a = - \dfrac{2}{3}\\
b \ne 1
\end{array} \right.\\
\Leftrightarrow y = - \dfrac{2}{3}x + b\left( {b \ne 1} \right)\\
M\left( {4;1} \right) \in y = - \dfrac{2}{3}x + b\\
\Leftrightarrow 1 = - \dfrac{2}{3}.4 + b\\
\Leftrightarrow b = 1 + \dfrac{8}{3} = \dfrac{{11}}{3}\left( {tmdk} \right)\\
Vậy\,a = - \dfrac{2}{3};b = \dfrac{{11}}{3}\\
f)y = \dfrac{{x - 2}}{3} = \dfrac{x}{3} - \dfrac{2}{3}\\
y = a.x + b \bot y = \dfrac{x}{3} - \dfrac{2}{3}\\
\Leftrightarrow a.\dfrac{1}{3} = - 1\\
\Leftrightarrow a = - 3\\
\Leftrightarrow y = - 3x + b\\
K\left( {2;2} \right) \in y = - 3x + b\\
\Leftrightarrow 2 = - 3.2 + b\\
\Leftrightarrow b = 8\\
Vậy\,a = - 3;b = 8
\end{array}$
