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