\[\boxed{\mathbf{1377.еуроки - ответы\ на\ пятёрку}}\]
\[\mathbf{Дано:}\]
\[A\left( x_{1};y_{1} \right);\]
\[B\left( x_{2};y_{2} \right);\]
\[C(x;y) \in AB;\]
\[AC\ :CB = \lambda.\]
\[\mathbf{Доказать:}\]
\[x = \frac{x_{1} + x_{2}}{1 + \lambda};\]
\[y = \frac{y + \lambda y}{1 + \lambda}.\]
\[\mathbf{Доказательство.}\]
\[1)\ \frac{\text{AC}}{\text{CB}} = \lambda:\]
\[\overrightarrow{\text{AC}}\left\{ x - x_{1};y - y_{1} \right\} =\]
\[= \lambda\overrightarrow{\text{CB}}\left\{ \lambda\left( x_{2} - x \right);\left( y_{2} - y \right) \right\}.\]
\[2)\left\{ \begin{matrix} x - x_{1} = \lambda\left( x_{2} - x \right) \\ y - y_{1} = \lambda\left( y_{2} - y \right) \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} x + \lambda x = x_{1} + \lambda x_{2} \\ y + \lambda y = y_{1} + \lambda y_{2} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = \frac{x_{1} + \lambda x_{2}}{1 + \lambda} \\ y = \frac{y_{1} + \lambda y_{2}}{1 + \lambda} \\ \end{matrix} \right.\ \]
\[Что\ и\ требовалось\ доказать.\]