\[y = kx;\ \ (2;2)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]
\[y = kx + b;\ \ (0;\ - 3)\]
\[Получаем\ систему\ уравнений:\]
\[\left\{ \begin{matrix}
y = x\ \ \ \ \ \ \ \ \\
y = x - 3 \\
\end{matrix} \right.\ \]
\[б)\ y = kx + b;\ \ (0;2)\]
\[Получаем\ систему\ уравнений:\]
\[\left\{ \begin{matrix}
y = x + 2 \\
y = - 2\ \ \ \ \\
\end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[в)\ y = kx + b;\ \ \ ( - 4;0);\ \ \ (4;2)\]
\[\left\{ \begin{matrix}
- 4k + b = 0 \\
4k + b = 2\ \ \ \ \\
\end{matrix} + \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix}
2b = 2\ \ \ \ \\
k = \frac{2 - b}{4} \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix}
b = 1 \\
k = \frac{1}{4} \\
\end{matrix} \right.\ \ \ \rightarrow y = \frac{1}{4}x + 1\]
\[y = kx + b;\ \ (2;0);\ ( - 2; - 1)\]
\[\left\{ \begin{matrix}
2k + b = 0\ \ \ \ \ \ \ \\
- 2k + b = - 1 \\
\end{matrix} + \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix}
2b = - 1\ \ \ \ \ \\
k = \frac{- 1 - b}{- 2} \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix}
b = - \frac{1}{2} \\
k = \frac{1}{4}\text{\ \ \ \ } \\
\end{matrix} \right.\ \ \rightarrow \ y = \frac{1}{4}x - \frac{1}{2}\]
\[Получаем\ систему\ уравнений:\]
\[\left\{ \begin{matrix}
y = \frac{1}{4}x + 1 \\
y = \frac{1}{4}x - \frac{1}{2} \\
\end{matrix} \right.\ \text{\ \ }\]