ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 114

\[\boxed{\text{115.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \ y = kx + b;\ \ \]

\[1)\ A(2;1);\ \ B( - 1;3)\]

\[1 = 2k + b;\ \ 3 = - k + b\]

\[b = 1 - 2k;\ \ \ b = 3 + k\]

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

\[- 3k = 2\]

\[k = - \frac{2}{3}.\]

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

\[Уравнение\ прямой:\]

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

\[(В\ учебнике\ неверный\ ответ).\]

\[2)\ A(1;5);\ \ B( - 1;2)\]

\[5 = k + b;\ \ \ \ 2 = - k + b\]

\[b = 5 - k;\ \ \ \ b = 2 + k\]

\[5 - k = 2 + k\]

\[- 2k = - 3\]

\[k = 1,5.\]

\[b = 2 + 1,5 = 3,5.\]

\[Уравнение\ прямой:\]

\[y = 1,5x + 3,5.\]

\[\textbf{б)}\ y = \frac{k}{x};\]

\[1)\ A(1;\ - 2)\]

\[- 2 = \frac{k}{1}\]

\[k = - 2.\]

\[Уравнение:\]

\[y = - \frac{2}{x}.\]

\[2)\ B(3;1)\]

\[1 = \frac{k}{3}\]

\[k = .\]

\[Уравнение:\]

\[y = \frac{3}{x}.\]

\[\textbf{в)}\ y = \sqrt{\text{kx}}\]

\[1)\ A(1;2)\]

\[2 = \sqrt{1k}\]

\[4 = k\]

\[k = 4.\]

\[Уравнение:\]

\[y = \sqrt{4x}.\]

\[2)\ B(1;3)\]

\[3 = \sqrt{1k}\]

\[k = 9.\]

\[Уравнение:\]

\[y = \sqrt{9x}.\]