\[\boxed{\mathbf{995.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Дано:}\]
\[M_{1}( - 2;4);\]
\[M_{2}(6;8);\]
\[X(x;0);\]
\[M_{1}X = M_{2}\text{X.}\]
\[\mathbf{Найти:}\]
\[XS - ?\]
\[\mathbf{Решение.}\]
\[1)\ M_{1}X =\]
\[= \sqrt{( - 2 - x)^{2} + (4 - 0)^{2}} =\]
\[= \sqrt{( - 2 - x)^{2} + 16};\]
\[M_{2}X = \sqrt{(6 - x)^{2} + (8 - 0)^{2}} =\]
\[= \sqrt{(6 - x)^{2} + 64}.\]
\[2)\ ( - 2 - x)^{2} + 16 =\]
\[= (6 - x)^{2} + 64\]
\[4 + 4x + x^{2} + 16 =\]
\[= 36 - 12x + x^{2} + 64.\]
\[16x = 80\]
\[x = 5.\]
\[X(5;0).\]
\(Ответ:X(5;0).\)
\[\boxed{\mathbf{995.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ \mathbf{задачи:}\]
\[\mathbf{Дано:}\]
\[A_{1}B_{1}C_{1}D_{1} - произвольный\ \]
\[четырехугольник;\]
\(A;B;C;D - середины\) \(сторон;\ \)
\[( \bullet )O - произвольная.\]
\[\mathbf{Доказать:}\]
\[\overrightarrow{\text{OA}} + \overrightarrow{\text{OC}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{OD}}.\]
\[\mathbf{Доказательство.}\]
\[1)\ BC - средняя\ линия\ \]
\[\mathrm{\Delta}B_{1}C_{1}D_{1}:\]
\[BC = \frac{1}{2}B_{1}D_{1}\ \ и\ BC \parallel B_{1}D_{1}.\]
\[Получаем:\]
\[BB_{1} = BC_{1};\ \ \ C_{1}C = CD_{1}.\]
\[2)\ AD - средняя\ линия\ \]
\[\mathrm{\Delta}A_{1}BD_{1}:\]
\[AD = \frac{1}{2}B_{1}D_{1}\ \ и\ \ AD \parallel B_{1}D_{1}.\]
\[Получаем:\]
\[AA_{1} = AB_{1};\ \ \ A_{1}D = DD_{1}.\]
\[3)\ BC \parallel B_{1}D_{1};\ \ AD \parallel B_{1}D_{1} \Longrightarrow\]
\[\Longrightarrow BC \parallel AD.\]
\[BC = \frac{1}{2}B_{1}D_{1};\ \ AD = \frac{1}{2}B_{1}D_{1} \Longrightarrow\]
\[\Longrightarrow \ BC = AD.\]
\[По\ определению\ равенства\ \]
\[векторов:\]
\[\overrightarrow{\text{BC}} = \overrightarrow{\text{AD}}.\]
\[4)\ \overrightarrow{\text{OC}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{BC}}\]
\[\overrightarrow{\text{BC}} = \overrightarrow{\text{OC}} - \overrightarrow{\text{OB}}\]
\[\overrightarrow{\text{OD}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{AD}} = \overrightarrow{\text{OA}} + \overrightarrow{\text{BC}}\]
\[\overrightarrow{\text{BC}} = \overrightarrow{\text{OD}} - \overrightarrow{\text{OA}}\]
\[Следовательно:\]
\[\overrightarrow{\text{OC}} - \overrightarrow{\text{OB}} = \overrightarrow{\text{OD}} - \overrightarrow{\text{OA}}.\]
\[Получаем:\]
\[\overrightarrow{\text{OA}} + \overrightarrow{\text{OC}} = \overrightarrow{\text{OB}} + \overrightarrow{\text{OD}}.\]
\[\mathbf{Что\ и\ требовалось\ доказать.}\]