ГДЗ по геометрии 9 класс Атанасян Задание 1003

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Год:2020-2021-2022
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Задание 1003

\[\boxed{\mathbf{1003.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

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

\[A( - 7;5);\]

\[B(3; - 1);\]

\[C(5;3).\]

\[\mathbf{Найти:}\]

\[уравнение\ прямых\]

\[\textbf{а)}\ AB;BC;AC;\]

\[\textbf{б)}\ серединных\ \]

\[перпендикуляров;\]

\[\textbf{в)}\ средних\ линий.\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ 1)\ AB:\ \]

\[\left\{ \begin{matrix} - 7a + 5b + c = 0 \\ 3a - b + c = 0\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 7a + 15a + 5c + c = 0 \\ b = 3a + c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 8a = - 6c\ \ \\ b = 3a + c \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a = - \frac{3}{4}c \\ b = - \frac{5}{4}c \\ \end{matrix} \right.\ \]

\[\left. \ - \frac{3}{4}cx - \frac{5}{4}cy + c = 0\ \ \ \ \ \right|\left( - \frac{4}{c} \right)\]

\[3x + 5y - 4 = 0.\]

\[2)\ BC:\ \]

\[\left\{ \begin{matrix} 3a - b + c = 0\ \ \\ 5a + 3b + c = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b = 3a + c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5a + 9a + 3c + c = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b = 3a + c \\ a = - \frac{2}{7}\text{c\ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} b = \frac{1}{7}\text{c\ \ \ \ } \\ a = - \frac{2}{7}c \\ \end{matrix} \right.\ \]

\[\left. \ - \frac{2}{7}cx + \frac{1}{7}cy + c = 0\ \ \ \ \ \ \right|\left( - \frac{7}{c} \right)\]

\[2x - y - 7 = 0.\]

\[3)\ AC:\ \]

\[\left\{ \begin{matrix} - 7a + 5b + c = 0 \\ 5a + 3b + c = 0\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 21a - 15b - 3c = 0 \\ 25a + 15b + 5c = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 46a = - 2c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 21a - 15b - 3c = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a = - \frac{1}{23}c \\ b = - \frac{6}{23}c \\ \end{matrix} \right.\ \]

\[x + 6y - 23 = 0.\]

\[4)\ MN:\ \]

\[\left\{ \begin{matrix} - 2a + 2b + c = 0 \\ 4a + b + c = 0\ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2a = 2b + c \\ b = - 4a - c \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} a = - \frac{1}{10}c \\ b = - \frac{6}{10}c \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left. \ - \frac{1}{10}cx - \frac{6}{10}cy + c = 0\ \ \right| \cdot \left( - \frac{10}{c} \right)\]

\[x + 6y - 10 = 0.\]

\[5)\ NK:\ \]

\[\left\{ \begin{matrix} 4a + b + c = 0\ \ \ \\ - a + 4b + c = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b = - 4a - c \\ a = 4b + c\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} b = - 16b - 4c - c \\ a = 4b + c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 17b = - 5c \\ a = 4b + c \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} b = - \frac{5}{17}c \\ a = - \frac{3}{17}c \\ \end{matrix} \right.\ \]

\[\left. \ - \frac{3}{17}cx - \frac{5}{17}cy + c = 0\ \ \ \ \right| \cdot \left( - \frac{17}{c} \right)\]

\[3x + 5y - 17 = 0.\]

\[5)\ MK:\ \]

\[\left\{ \begin{matrix} - 2a + 2b + c = 0 \\ - a + 4b + c = 0\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 2b - 2c + 2b + c = 0 \\ a = 4b + c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 6b = c\ \ \ \ \\ a = 4b + c \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} b = - \frac{1}{6}c \\ a = \frac{1}{3}\text{c\ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left. \ \frac{1}{3}cx - \frac{1}{6}cy + c = 0\ \ \ \ \right| \cdot \left( \frac{6}{c} \right)\]

\[2x - y + 6 = 0.\]

\[\textbf{в)}\ 1)\ l_{1}\bot AB.\ \ \ \]

\[AB:\]

\[3x + 5y - 4 = 0.\]

\[l_{1}:\]

\[ax + by + c = 0.\]

\[5y = 4 - 3x\]

\[y = - \frac{3}{5}x + \frac{4}{5} \Longrightarrow k_{1} = - \frac{3}{5}.\]

\[by = - ax - c\]

\[y = - \frac{a}{b}x - \frac{c}{b} \Longrightarrow k_{2} = - \frac{a}{b}.\]

\[2)\ k_{1} \bullet k_{2} = - 1 \Longrightarrow - \frac{3}{5} \bullet \left( - \frac{a}{b} \right) =\]

\[= - 1 \Longrightarrow \frac{3a}{5b} = - 1;\]

\[3a = - 5b \Longrightarrow a = 5;b = - 3 \Longrightarrow\]

\[\Longrightarrow 5x - 3y + c = 0.\]

\[3)\ M \in l_{1}:\ \ \ \]

\[5( - 2) - 3 \bullet 2 + c = 0\]

\[c = 16.\]

\[l_{1}:\]

\[5x - 3y + 16 = 0.\]

\[4)\ l_{2}\bot AC.\ \ \ \]

\[AC:\]

\[x + 6y - 23 = 0.\]

\[l_{2}:\]

\[ax + by + c = 0.\]

\[6y = 23 - x\]

\[y = - \frac{1}{6}x + \frac{23}{6} \Longrightarrow k_{1} = - \frac{1}{6}.\]

\[by = - ax - c\]

\[y = - \frac{a}{b}x - \frac{c}{b} \Longrightarrow k_{2} = - \frac{a}{b}.\]

\[5)\ k_{1} \bullet k_{2} = - 1 \Longrightarrow - \frac{1}{6} \bullet \left( - \frac{a}{b} \right) =\]

\[= - 1 \Longrightarrow \frac{a}{6b} = - 1;\]

\[a = - 6b \Longrightarrow a = 6;b = - 1 \Longrightarrow\]

\[\Longrightarrow 6x - y + c = 0.\]

\[6)\ K \in l_{2}:\ \ \ \]

\[6( - 1) - 4 + c = 0 \Longrightarrow c = 10.\]

\[l_{2}:\]

\[6x - y + 10 = 0.\]

\[7)\ l_{3}\bot BC.\ \ \ \]

\[BC:\]

\[2x - y - 7 = 0.\]

\[l_{3}:\]

\[ax + by + c = 0.\]

\[- y = 7 - 2x\]

\[y = 2x - 7 \Longrightarrow k_{1} = 2.\]

\[by = - ax - c\]

\[y = - \frac{a}{b}x - \frac{c}{b} \Longrightarrow k_{2} = - \frac{a}{b}.\]

\[8)\ k_{1} \bullet k_{2} = - 1 \Longrightarrow 2 \bullet \left( - \frac{a}{b} \right) =\]

\[= - 1 \Longrightarrow \frac{2a}{b} = 1;\]

\[b = 2a \Longrightarrow a = 1;\]

\[b = 2 \Longrightarrow x + 2y + c = 0.\]

\[9)\ N \in l_{3}:\ \ \ \]

\[4 + 2 + c = 0 \Longrightarrow c = - 6.\]

\[l_{3}:\]

\[x + 2y - 6 = 0.\]

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