ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 1191

\[\boxed{\text{1191.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ \text{A\ }(1;2)\ и\ \ \text{B\ }( - 2;3):\]

\[\left\{ \begin{matrix} 2 = k + b\ \ \ | \cdot ( - 1) \\ 3 = - 2k + b\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} - 2 = - k - b \\ 3 = - 2k + b \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 1 = - 3k\ \\ b = 2 - k \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} k = - \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ b = 2 + \frac{1}{3} = 2\frac{1}{3} \\ \end{matrix} \right.\ \Longrightarrow \ y =\]

\[= - \frac{1}{3}x + 2\frac{1}{3}.\]

\[\textbf{б)}\ \text{M\ }( - 5;0)\ \ и\ \ \ K\ (2;\ - 1)\]

\[\left\{ \begin{matrix} 0 = - 5k + b\ \ | \cdot ( - 1) \\ - 1 = 2k + b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 5k - b = 0\ \ \\ 2k + b = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 7k = - 1 \\ b = 5k\ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} k = - \frac{1}{7} \\ b = - \frac{5}{7} \\ \end{matrix} \right.\ \]

\[\Longrightarrow \ \ y = - \frac{1}{7}x - \frac{5}{7}.\]