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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 974
Задание 974
\[\boxed{\mathbf{974.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[A( - 2; - 2);\]
\[B( - 3;1);\]
\[C(7;7);D(3;1)\]
\[MN - средняя\ линия.\]
\[\mathbf{Написать:}\]
\[уравнения\ прямых\]
\[\textbf{а)}\ \text{AC\ }и\ BD;\]
\[\textbf{б)}\ MN.\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ A( - 2; - 2):\]
\[ax + by + c = 0;\]
\[a \bullet ( - 2) + b \bullet ( - 2) + c = 0;\]
\[- 2a - 2b + c = 0;\ \]
\[- 2a = 2b - c;\]
\[a = \frac{c - 2b}{2} = \frac{c}{2} - b.\]
\[C(7;7):\]
\[ax + by + c = 0;\]
\(7a + 7b + c = 0;7a = - 7b - c;\)
\[a = \frac{- 7b - c}{7} = - b - \frac{c}{7};\]
\[- b - \frac{c}{7} = \frac{c}{2} - b\]
\[\left\{ \begin{matrix} c = 0 \\ a = - b \\ \end{matrix} \right.\ .\]
\[2)\ Уравнение\ прямой\ AC:\]
\[\left. \ ax - ay + 0 = 0\ \ \ \ \ \ \ \ \right|:a\]
\[x - y = 0.\]
\[3)\ B( - 3;1):\]
\[ax + by + c = 0;\]
\[- 3a + b + c = 0;\]
\[a = \frac{b + c}{3}.\]
\[D(3;1):\]
\[ax + by + c = 0;\]
\[3a + b + c = 0;\]
\[b = - 3a - c;\]
\[a = \frac{- 3a - c + c}{3} = - a\]
\[a = 0 \Longrightarrow b = - c.\]
\[4)\ Уравнение\ прямой\ BD:\]
\[ax + by + c = 0\]
\[- cy + c = 0\ \ \ \ |\ :c\]
\[- y + 1 = 0.\]
\[3)\ N(5;4):\]
\[5a + 4b + c = 0\]
\[a = \frac{- 4b - c}{5}.\]
\[M( - 2,5; - 0,5):\ \]
\[\left. \ - \frac{5}{2}a - \frac{1}{2}b + c = 0 \right| \bullet 2\]
\[- 5a - b + 2c = 0;b = 2c - 5a\]
\[b = 2c - ( - 4b + c) =\]
\[= 2c + 4b + c + 4b\]
\[- 3b = 3c\]
\[b = - c \Longrightarrow a = \frac{3}{5}\text{c.}\]
\[4)\ Уравнение\ прямой\ MN:\]
\[\left. \ \frac{3}{5}cx - cy + c = 0 \right| \bullet \frac{5}{c}\]
\[3x - 5y + 5 = 0.\]
\[Ответ:\ а)\ x - y = 0\ и\ \]
\[- y + 1 = 0;\ \]
\[\textbf{б)}\ 3x - 5y + 5 = 0.\]
\[\boxed{\mathbf{974}\mathbf{.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\mathbf{\ задачи:}\]
\[\mathbf{Дано:}\]
\[\mathrm{\Delta}ABC;\]
\[AA_{1};BB_{1};CC_{1} - медианы.\]
\[\overrightarrow{a} = \overrightarrow{\text{AC}};\ \ \]
\[\overrightarrow{b} = \overrightarrow{\text{AB}}.\]
\[Выразить:\]
\[\overrightarrow{AA_{1}};\ \ \overrightarrow{BB_{1}\ };\ \overrightarrow{CC_{1}}\mathbf{.}\]
\[\mathbf{Решение.}\]
\[1)\ \overrightarrow{BB_{1}} = \overrightarrow{\text{BA}} + \ \overrightarrow{AB_{1}} =\]
\[= - \overrightarrow{\text{AB}} + \frac{1}{2}\overrightarrow{\text{AC}} = - \overrightarrow{b} + \frac{1}{2}\overrightarrow{a}.\]
\[2)\ \overrightarrow{CC_{1}} = \overrightarrow{\text{CB}} + \overrightarrow{BC_{1}} =\]
\[= \overrightarrow{\text{CA}} + \overrightarrow{AC_{1}} = - \overrightarrow{\text{AC}} + \frac{1}{2}\overrightarrow{\text{AB}} =\]
\[= \overrightarrow{a} + \frac{1}{2}\overrightarrow{b}.\]
\[3)\ \overrightarrow{\text{BC}} = \overrightarrow{\text{BA}} + \overrightarrow{\text{AC}} =\]
\[= - \overrightarrow{\text{AB}} + \overrightarrow{\text{AC}} = - \overrightarrow{b} + \overrightarrow{a}.\]
\[4)\ \overrightarrow{AA_{1}} = \overrightarrow{\text{AB}} + \overrightarrow{BA_{1}} =\]
\[= \overrightarrow{\text{AB}} + \frac{1}{2}\overrightarrow{\text{BC}} = \overrightarrow{b} + \frac{1}{2} \bullet \left( \overrightarrow{a} - \overrightarrow{b} \right) =\]
\[= \overrightarrow{b} + \frac{1}{2}\overrightarrow{a} - \frac{1}{2}\overrightarrow{b} =\]
\[= \frac{1}{2}\overrightarrow{b} + \frac{1}{2}\overrightarrow{a}.\]
