ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1458

\[\boxed{\mathbf{1458}\mathbf{.}}\]

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

\[1)\ A - пересечение\ с\ осью\ \text{Ox\ }\]

\[(y = 0):\]

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

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

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

\[B - пересечение\ с\ осью\ \text{Oy\ }\]

\[(x = 0):\]

\[y = - \frac{3}{4} \bullet 0 + 2 = 0 + 2 = 2.\]

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

\[2)\ Длина\ отрезка\ AB:\]

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

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

\[Ответ:\ \ 3\frac{1}{3}.\]

\[3)\ Проходит\ через\ точку\ O(0;\ 0):\]

\[k_{1} = - \frac{1}{k} = - \left( - \frac{4}{3} \right) = \frac{4}{3};\]

\[0 = \frac{4}{3} \bullet 0 + b_{1}\]

\[b_{1} = 0;\]

\[y_{1} = \frac{4}{3}\text{x.}\]

\[C - пересечение\ прямых:\]

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

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

\[25x = 24\]

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

\[y = \frac{4}{3} \bullet \frac{24}{25} = \frac{4 \bullet 8}{25} = \frac{32}{25}.\]

\[Длина\ отрезка\ OC:\]

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

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

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

\[Ответ:\ \ \frac{8}{5}.\]