\[\boxed{\mathbf{1010.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[точки\ A\ и\ B;\]
\[\textbf{а)}\ 2AM^{2} - BM^{2} = 2AB^{2};\]
\[\textbf{б)}\ 2AM^{2} + 2BM^{2} = 6AB^{2}.\]
\[\mathbf{Найти:}\]
\[множество\ точек\ \text{M.}\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ Введем\ систему\ \]
\[координат:\]
\[AB \in OX;A(0;0);B(a;0);\]
\[M(x;y).\]
\[2)\ AM^{2} = x^{2} + y^{2};\ \ \ AB^{2} = a^{2};\]
\[BM^{2} = (a - x)^{2} + y^{2}.\]
\[3)\ 2\left( x^{2} + y^{2} \right) - \left( (a - x)^{2} + y^{2} \right) =\]
\[= 2a^{2}\]
\[2x^{2} + 2y^{2} - a^{2} + 2ax - x^{2} - y^{2} =\]
\[= 2a^{2}\]
\[x^{2} + y^{2} + 2ax = 3a^{2}\]
\[(x + a)^{2} + y^{2} = 4a^{2}\text{.\ }\]
\[3)\ Множество\ точек\ M:\ \]
\[окружность\ с\ центром\ в\ точке\ \]
\[( - a;0)\ и\ R = 2a.\]
\[\textbf{б)}\ 1)\ Введем\ систему\ \]
\[координат:\]
\[AB \in OX;A(0;0);B(a;0);\]
\[M(x;y).\]
\[2)\ AM^{2} = x^{2} + y^{2};\ \ \ AB^{2} = a^{2};\]
\[BM^{2} = (a - x)^{2} + y^{2}.\]
\[4x^{2} + 4y^{2} - 4ax = 4a^{2}\]
\[\left( 4x^{2} - 4ax + a^{2} \right) - a^{2} + 4y^{2} =\]
\[= 4a^{2}\]
\[(2x - a)^{2} + 4y^{2} = 5a^{2}\]
\[\left. \ 4\left( x - \frac{a}{2} \right)^{2} + 4y^{2} = 5a^{2}\text{\ \ \ \ \ } \right|:4\]
\[\ \left( x - \frac{a}{2} \right)^{2} + y^{2} = \frac{5a^{2}}{4}.\]
\[3)\ Множество\ точек\ M:\]
\[окружность\ с\ центром\ в\ точке\ \]
\[\left( \frac{a}{2};0 \right)\ и\ R = \frac{\sqrt{5}}{2}\text{a.}\]
\[\mathbf{Глава\ 11.\ Соотношения\ между\ сторонами\ и\ углами\ треугольника.}\]
\[\mathbf{Скалярное\ произведение\ векторов}\]
\[\mathbf{Параграф\ }1\mathbf{.\ Синус,\ косинус,\ тангенс,\ котангенс\ угла}\]
\[\boxed{\mathbf{1010.еуроки - ответы\ на\ пятёрку}}\]
\[\overrightarrow{a} - \overrightarrow{b} = \ ?\]
\[\textbf{а)}\ \overrightarrow{a}\ \left\{ 5;3 \right\};\ \overrightarrow{b}\ \left\{ 2;1 \right\} \Longrightarrow\]
\[\Longrightarrow \ \overrightarrow{a} - \overrightarrow{b}\ \left\{ 3;2 \right\}.\]
\[\textbf{б)}\ \overrightarrow{a}\ \left\{ 3;2 \right\};\ \overrightarrow{b}\ \left\{ - 3;2 \right\} \Longrightarrow \ \]
\[\Longrightarrow \overrightarrow{a} - \overrightarrow{b}\ \left\{ 6;0 \right\}.\]
\[\textbf{в)}\ \overrightarrow{a}\ \left\{ 3;6 \right\};\ \overrightarrow{b}\ \left\{ 4; - 3 \right\} \Longrightarrow\]
\[\Longrightarrow \ \overrightarrow{a} - \overrightarrow{b}\ \left\{ - 1;9 \right\}.\]
\[\textbf{г)}\ \overrightarrow{a}\ \left\{ - 5; - 6 \right\};\ \overrightarrow{b}\ \left\{ 2; - 4 \right\} \Longrightarrow \ \]
\[\Longrightarrow \overrightarrow{a} - \overrightarrow{b}\ \left\{ - 7; - 2 \right\}.\]