\[\boxed{\mathbf{973.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[A(4;6);\ \]
\[B( - 4;0);\]
\[C( - 1; - 4);\]
\[CM - медиана\ \mathrm{\Delta}\text{ABC.}\]
\[\mathbf{Написать:}\]
\[уравнение\ прямой\ \text{CM.}\]
\[\mathbf{Решение.}\]
\[2)\ M(0;3):\]
\[ax + by + c = 0;\]
\[0 \bullet a + 3b + c = 0;\]
\[3b + c = 0;\ \ \]
\[b = - \frac{c}{3}.\]
\[C( - 1; - 4):\]
\[ax + by + c = 0;\]
\[a( - 1) + b( - 4) + c = 0;\]
\[- a - 4b + c =\]
\[= - a - 4 \bullet \left( - \frac{c}{3} \right) + c;\]
\[- a = - c - \frac{4c}{3};\ \ \]
\[a = \frac{7c}{3}.\]
\[3)\ Уравнение\ прямой\ CM:\]
\[ax + by + c = 0\]
\[\frac{7c}{3}x - \left. \ \frac{c}{3}y + c = 0\ \ \ \ \ \ \ \ \ \ \right| \bullet \frac{3}{c}\]
\[7x - y + 3 = 0.\]
\[Ответ:\ 7x - y + 3 = 0.\]
\[\boxed{\mathbf{973}\mathbf{.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[ABCD - четырехугольник;\]
\[AM = MC;\ \ \]
\[BN = ND.\]
\[\mathbf{Доказать:}\]
\[\overrightarrow{\text{MN}} = \frac{1}{2} \bullet \left( \overrightarrow{\text{AD}} + \overrightarrow{\text{CB}} \right).\]
\[\mathbf{Доказательство.}\]
\[1)\ По\ правилу\ \]
\[многоугольников:\]
\[\ \overrightarrow{\text{MN}} = \overrightarrow{\text{MA}} + \overrightarrow{\text{AD}} + \overrightarrow{\text{DN}}\]
\[+\]
\[\]
\[Отсюда:\]
\[\overrightarrow{\text{MA}} + \overrightarrow{\text{MC}} = \overrightarrow{0}\ \ \ и\ \ \ \overrightarrow{\text{DN}} + \overrightarrow{\text{BN}} = \overrightarrow{0};\]
\[так\ как:\]
\[AM = MC\ \ \ и\ \ \overrightarrow{\text{MA}} \nearrow \swarrow \overrightarrow{\text{MC}};\]
\[BN = DN\ \ \ и\ \ \overrightarrow{\text{DN}} \nearrow \swarrow \overrightarrow{\text{BN}}.\]
\[Следовательно:\]
\[2\overrightarrow{\text{MN}} = \overrightarrow{\text{AD}} + \overrightarrow{\text{CB}};\]
\[\overrightarrow{\text{MN}} = \frac{1}{2} \bullet \left( \overrightarrow{\text{AD}} + \overrightarrow{\text{CB}} \right).\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]