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МатематикаРешебник по геометрии 7 класс Атанасян ФГОС Задание 999
Задание 999
\[\boxed{\mathbf{999.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[A( - 4;4);B( - 5;1);\]
\[C(x;y);D( - 1;5).\]
\[\mathbf{Найти:}\]
\[C(x;y) - ?\]
\[\mathbf{Решение.}\]
\[1)\ AB =\]
\[= \sqrt{( - 4 + 5)^{2} + (4 - 1)^{2}} = \sqrt{10};\]
\[AD = \sqrt{( - 4 + 1)^{2} + (4 - 5)^{2}} =\]
\[= \sqrt{10};\]
\[BC = \sqrt{( - 5 - x)^{2} + (1 - y)^{2}};\]
\[CD = \sqrt{(x + 1)^{2} + (y - 5)^{2}}.\]
\[2)\ ABCD - параллелограмм:\]
\[AB = CD\ и\ \]
\[BC = AD\ (по\ свойству).\]
\[\left\{ \begin{matrix} (x + 1)^{2} + (y - 5)^{2} = 10\ \ \ \\ ( - 5 - x)^{2} + (1 - y)^{2} = 10 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + 2x + 1 + y^{2} - 10y + 25 = 10 \\ 25 + 10x + x^{2} + 1 - 2y + y^{2} = 10 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 8x + 8y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 1)^{2} + (y - 5)^{2} = 10 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (1 - y)^{2} + (y - 5)^{2} = 10 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 6y + 8 = 0 \\ \end{matrix} \right.\ \]
\[3)\left\{ \begin{matrix} y = 2 \\ x = - 2 \\ \end{matrix} \right.\ \ или\ \left\{ \begin{matrix} y = 4 \\ x = - 4 \\ \end{matrix} \right.\ \]
\[( - 4;4) - уже\ есть\ (точка\ A);\]
\[C( - 2;2).\]
\[Ответ:C( - 2;2).\]
\[\boxed{\mathbf{999.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[BD \cap AC = O;\]
\[AM = MO;\]
\[M \in AO.\]
\[\mathbf{Найти:}\]
\[k - ?\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ \overrightarrow{\text{AC}} = k\overrightarrow{\text{AO}};\ \overrightarrow{\text{AC}} \uparrow \uparrow \overrightarrow{\text{AO}}:\]
\[k > 0 \Longrightarrow \left| \overrightarrow{\text{AC}} \right| = 2\left| \overrightarrow{\text{AO}} \right| \Longrightarrow\]
\[\Longrightarrow k = 2;\]
\[\textbf{б)}\ \overrightarrow{\text{BO}} = k\overrightarrow{\text{BD}};\ \overrightarrow{\text{BO}} \uparrow \uparrow \overrightarrow{\text{BD}}:\]
\[k > 0 \Longrightarrow \left| \overrightarrow{\text{BO}} \right| = \frac{1}{2}\left| \overrightarrow{\text{BD}} \right| \Longrightarrow\]
\[\Longrightarrow k = \frac{1}{2};\]
\[\textbf{в)}\ \overrightarrow{\text{OC}} = k\overrightarrow{\text{CA}};\ \overrightarrow{\text{OC}} \uparrow \downarrow \overrightarrow{\text{CA}}:\]
\[k < 0 \Longrightarrow \left| \overrightarrow{\text{OC}} \right| = \frac{1}{2}\left| \overrightarrow{\text{CA}} \right| \Longrightarrow\]
\[\Longrightarrow k = - \frac{1}{2};\]
\[\textbf{г)}\ \overrightarrow{\text{AB}} = k\overrightarrow{\text{DC}};\ \overrightarrow{\text{AB}} \uparrow \uparrow \overrightarrow{\text{DC}}:\]
\[k > 0 \Longrightarrow \left| \overrightarrow{\text{AB}} \right| = 1\left| \overrightarrow{\text{DC}} \right| \Longrightarrow\]
\[\Longrightarrow k = 1;\]
\[\textbf{д)}\ \overrightarrow{\text{BC}} = k\overrightarrow{\text{DA}};\ \overrightarrow{\text{BC}} \uparrow \downarrow \overrightarrow{\text{DA}}:\]
\[k < 0 \Longrightarrow \left| \overrightarrow{\text{BC}} \right| = 1\left| \overrightarrow{\text{DA}} \right| \Longrightarrow\]
\[\Longrightarrow k = - 1;\]
\[\textbf{е)}\ \overrightarrow{\text{AM}} = k\overrightarrow{\text{CA}};\ \overrightarrow{\text{AM}} \uparrow \downarrow \overrightarrow{\text{CA}}:\]
\[k < 0 \Longrightarrow \left| \overrightarrow{\text{AM}} \right| = \frac{1}{4}\left| \overrightarrow{\text{CA}} \right| \Longrightarrow\]
\[\Longrightarrow k = - \frac{1}{4};\]
\[\textbf{ж)}\ \overrightarrow{\text{MC}} = k\overrightarrow{\text{AM}};\ \overrightarrow{\text{MC}} \uparrow \uparrow \overrightarrow{\text{AM}}:\]
\[k > 0 \Longrightarrow \left| \overrightarrow{\text{MC}} \right| = 3\left| \overrightarrow{\text{AM}} \right| \Longrightarrow\]
\[\Longrightarrow k = 3;\]
\[\textbf{з)}\ \overrightarrow{\text{AC}} = k\overrightarrow{\text{CM}};\ \overrightarrow{\text{AC}} \uparrow \downarrow \overrightarrow{\text{CM}}:\]
\[k < 0 \Longrightarrow \left| \overrightarrow{\text{AC}} \right| = \frac{4}{3}\left| \overrightarrow{\text{CM}} \right| \Longrightarrow\]
\[\Longrightarrow k = - \frac{4}{3};\]
\[\textbf{и)}\ \overrightarrow{\text{AO}} = k:\ \]
\[\overrightarrow{\text{AO}}\ и\ \overrightarrow{\text{BD}} - неколлинеарны \Longrightarrow\]
\[\Longrightarrow их\ нельзя\ разложить.\]
