\[\boxed{\mathbf{937.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Дано:}\]
\[\text{A\ }(0;1);\]
\[\text{B\ }(5; - 3);\]
\[AB = BC;\]
\[BD = DC;\]
\[B \in AC;\ \]
\[D \in BC.\]
\[\mathbf{Найти:}\]
\[координаты\ C\ и\ \text{D.}\]
\[\mathbf{Решение.}\]
\[1)\ \left\{ \begin{matrix} x_{B} = \frac{x_{A} + x_{C}}{2} \\ y_{B} = \frac{y_{A} + y_{C}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} 5 = \frac{0 + x_{C}}{2}\text{\ \ \ \ } \\ - 3 = \frac{1 + y_{C}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} \frac{x_{C}}{2} = 5\ \ \ \ \ \ \ \\ \frac{y_{C}}{2} = - 3,5 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x_{C} = 10\ \\ y_{C} = - 7 \\ \end{matrix} \Longrightarrow C\ (10; - 7). \right.\ \]
\[2)\ \left\{ \begin{matrix} x_{D} = \frac{x_{B} + x_{C}}{2} \\ y_{D} = \frac{y_{B} + y_{c}}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{D} = \frac{5 + 10}{2}\text{\ \ \ \ \ \ \ \ } \\ y_{D} = \frac{- 3 + ( - 7)}{2} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x_{D} = 7,5 \\ y_{D} = - 5 \\ \end{matrix}\ \Longrightarrow D\ (7,5;\ - 5). \right.\ \]
\(Ответ:C\ (10; - 7);D\ (7,5; - 5).\)