Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 681

\[z_{1} = x_{1} + y_{1}i;\text{\ \ }\]

\[z_{2} = x_{2} + y_{2}i;\]

\[z_{3} = x_{3} + y_{3}i.\]

\[1)\ Середина\ стороны\ z_{2}z_{3}:\]

\[x_{m} = \frac{x_{2} + x_{3}}{2};\]

\[y_{m} = \frac{y_{2} + y_{3}}{2}.\]

\[2)\ Точка\ пересечения\ медиан:\]

\[z_{1}z_{o}\ :z_{o}z_{m} = 2\ :1\]

\[\frac{x_{o} - x_{1}}{x_{m} - x_{o}} = 2;\text{\ \ \ }\]

\[\frac{y_{o} - y_{1}}{y_{m} - y_{o}} = 2.\]

\[x_{o} - x_{1} = 2x_{m} - 2x_{o}\text{\ \ \ }\]

\[3x_{o} = x_{1} + 2x_{m}\text{\ \ \ }\]

\[x_{o} = \frac{x_{1} + 2x_{m}}{3}\text{\ \ }\]

\[x_{o} = \frac{x_{1} + x_{2} + x_{3}}{3}.\]

\[y_{o} - y_{1} = 2y_{m} - 2y_{o}\]

\[3y_{o} = y_{1} + 2y_{m}\]

\[y_{o} = \frac{y_{1} + 2y_{m}}{3}\]

\[y_{o} = \frac{y_{1} + y_{2} + y_{3}}{3}.\]

\[z_{o} = x_{o} + y_{o}i =\]

\[= \frac{x_{1} + x_{2} + x_{3} + y_{1}i + y_{2}i + y_{3}i}{3} =\]

\[= \frac{\left( x_{1} + y_{1}i \right) + \left( x_{2} + y_{2}i \right) + \left( x_{3} + y_{3}i \right)}{3} =\]

\[z_{o} = \frac{z_{1} + z_{2} + z_{3}}{3}.\]

\[Ответ:\ \ \frac{1}{3}\left( z_{1} + z_{2} + z_{3} \right).\]