Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1455

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

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

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

\[3(b + 2) + b = 2\]

\[3b + 6 + b = 2\]

\[4b = - 4\]

\[b = - 1;\]

\[k = - 1 + 2 = 1.\]

\[Ответ:\ \ k = 1;\ \ b = - 1.\]

\[2)\ A(2;\ 1)\text{\ \ }и\ \ B(1;\ 2):\]

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

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

\[2(2 - b) + b = 1\]

\[4 - 2b + b = 1\]

\[- b = - 3\]

\[b = 3;\]

\[k = 2 - 3 = - 1.\]

\[Ответ:\ \ k = - 1;\ \ b = 3.\]

\[3)\ A(4;\ 2)\text{\ \ }и\ \ B( - 4;\ - 3):\]

\[\left\{ \begin{matrix} 2 = 4k + b\ \ \ \ \ \ \ \\ - 3 = - 4k + b \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} b = 2 - 4k \\ b = 4k - 3 \\ \end{matrix} \right.\ \]

\[2 - 4k = 4k - 3\]

\[- 8k = - 5\]

\[k = \frac{5}{8};\]

\[b = 2 - 4 \bullet \frac{5}{8} = 2 - \frac{5}{2} = - \frac{1}{2}.\]

\[Ответ:\ \ k = \frac{5}{8};\ \ b = - \frac{1}{2}.\]

\[4)\ A( - 2;\ - 2)\text{\ \ }и\ \ B(3;\ - 2):\]

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

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

\[2k - 2 = - 3k - 2\]

\[5k = 0\]

\[k = 0;\]

\[b = 2 \bullet 0 - 2 = - 2.\]

\[Ответ:\ \ k = 0;\ \ b = - 2.\]