ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 857

\[\boxed{\mathbf{857}\mathbf{.}}\]

\[y = kx + b\ \ k = tg\ a\]

\[1)\ a = \frac{\pi}{4},\ \ \ x_{0} = 2,\ \ \ y_{0} = - 3:\]

\[k = tg\ a = tg\frac{\pi}{4} = 1\]

\[- 3 = 2 + b\]

\[b = - 5.\]

\[Ответ:\ \ \ k = 1\ \ b = - 5.\]

\[2)\ a = \frac{\pi}{4},\ \ \ x_{0} = - 3,\ \ \ y_{0} = 2:\]

\[k = tg\ a = tg\frac{\pi}{4} = 1\]

\[2 = - 3 + b\]

\[b = 5.\]

\[Ответ:\ \ k = 1\ \ b = 5\]

\[3)\ a = - \frac{\pi}{3},\ \ \ x_{0} = 1,\ \ \ y_{0} = 1:\]

\[k = tg\ a = - tg\frac{\pi}{3} = - \sqrt{3}\]

\[1 = - \sqrt{3} + b\]

\[b = 1 + \sqrt{3}.\]

\[Ответ:\ \ k = - \sqrt{3}\ \ b = 1 + \sqrt{3}.\]

\[4)\ a = - \frac{\pi}{6},\ \ \ x_{0} = - 1,\ \ \ y_{0} = - 1:\]

\[k = tg\ a = - tg\frac{\pi}{6} = - \frac{\sqrt{3}}{3}\]

\[- 1 = \frac{\sqrt{3}}{3} + b\]

\[b = - 1 - \frac{\sqrt{3}}{3}.\]

\[Ответ:\ \ k = - \frac{\sqrt{3}}{3}\text{\ \ }\]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }b = - 1 - \frac{\sqrt{3}}{3}.\]