Решебник по геометрии 7 класс Атанасян ФГОС Задание 991

 

\[\boxed{\mathbf{991.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Дано:\]

\[M_{1}\left( x_{1};0 \right);\]

\[M_{2}\left( x_{2};0 \right).\]

\[Доказаь:\]

\[d = \left| x_{1} - x_{2} \right|.\]

\[Доказательство.\]

\[d = M_{1}M_{2} =\]

\[= \sqrt{\left( x_{1} - x_{2} \right)^{2} + (0 - 0)} =\]

\[= \sqrt{\left( x_{1} - x_{2} \right)^{2}} = \left| x_{1} - x_{2} \right|.\]

\[Что\ и\ требовалось\ доказать.\]

\[\boxed{\mathbf{991.еуроки - ответы\ на\ пятёрку}}\]

\[Рисунок\ по\ условию\ \mathbf{задачи:}\]

\[\mathbf{Дано:}\]

\[\mathrm{\Delta}MNP;\]

\[X \in MN;Y \in NP.\]

\[\frac{\text{MX}}{\text{XN}} = \frac{3}{2};\ \frac{\text{NY}}{\text{YP}} = \frac{3}{2};\]

\[\overrightarrow{a} = \overrightarrow{\text{NM}};\ \ \overrightarrow{b} = \overrightarrow{\text{NP}}.\]

\[Выразить:\]

\[\overrightarrow{\text{XY}}\ \ и\ \overrightarrow{\text{MP}}\ через\ \overrightarrow{a}\ и\ \overrightarrow{b}.\]

\[\mathbf{Решение.}\]

\[1)\ \frac{\text{MX}}{\text{XN}} = \frac{3}{2}:\]

\[\overrightarrow{\text{NM}} = 5\ частей;\]

\[\overrightarrow{\text{NX}} = \frac{2}{2}\overrightarrow{\text{NM}} = \frac{2}{5}\overrightarrow{a}.\]

\[2)\ \frac{\text{NY}}{\text{YP}} = \frac{3}{2}:\]

\[\overrightarrow{\text{NP}} = 5\ частей;\]

\[\overrightarrow{\text{NY}} = \frac{3}{5}\overrightarrow{\text{NP}} = \frac{3}{5}\overrightarrow{b}.\]

\[4)\ \overrightarrow{\text{XY}} = \overrightarrow{\text{XN}} + \overrightarrow{\text{NY}} =\]

\[= - \overrightarrow{\text{NX}} + \overrightarrow{\text{NY}} = - \frac{2}{5}\overrightarrow{a} + \frac{3}{5}\overrightarrow{b}.\]

\[4)\ \overrightarrow{\text{MP}} = \overrightarrow{\text{MN}} + \overrightarrow{\text{NP}} =\]

\[= - \overrightarrow{\text{NM}} + \overrightarrow{\text{NP}} = - \overrightarrow{a} + \overrightarrow{b}.\]

\[Ответ:\overrightarrow{\text{XY}} = - \frac{2}{5}\overrightarrow{a} + \frac{3}{5}\overrightarrow{b};\ \]

\[\ \overrightarrow{\text{MP}} = - \overrightarrow{a} + \overrightarrow{b}\text{.\ \ }\]