ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 998

\[y = - \frac{3}{4}x + 2.\]

\[1)\ Пересечение\ с\ осями\ \]

\[координат:\]

\[- \frac{3}{4}x + 2 = 0\]

\[- \frac{3}{4}x = - 2\]

\[x = \frac{8}{3};\ \ \ \]

\[y(0) = 2.\]

\[Ответ:\ \ A\left( \frac{8}{3};\ 0 \right);\ B(0;\ 2).\]

\[2)\ AB = \sqrt{\left( \frac{8}{3} \right)^{2} + 2^{2}} = \sqrt{\frac{64}{9} + 4} =\]

\[= \sqrt{\frac{64 + 36}{9}} = \sqrt{\frac{100}{9}} = \frac{10}{3} = 3\frac{1}{3}.\]

\[Ответ:\ \ 3\frac{1}{3}.\]

\[3)\ - \frac{3}{4}x + 2 = \frac{4}{3}x\ \ \ \ \ | \bullet 12\]

\[- 9x + 24 = 16x\]

\[25x = 24\]

\[x = \frac{24}{25};\ \ \ \]

\[y = \frac{4}{3} \bullet \frac{24}{25} = \frac{32}{25}.\]

\[OC = \sqrt{\left( \frac{24}{25} \right)^{2} + \left( \frac{32}{25} \right)^{2}} =\]

\[= \sqrt{\frac{576 + 1024}{25^{2}}} = \sqrt{\frac{1600}{25^{2}}} =\]

\[= \frac{40}{25} = \frac{8}{5}.\]

\[Ответ:\ \ \frac{8}{5}.\]