Решебник по геометрии 7 класс Атанасян ФГОС Задание 1047

 

\[\boxed{\mathbf{1047.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\ \]

\[\overrightarrow{a}\ и\ \overrightarrow{b}.\]

\[Найти:\]

\[x,\ при\ котором\ \overrightarrow{a}\bot\overrightarrow{b}.\]

\[Решение:\]

\[Если\ \overrightarrow{a}\bot\overrightarrow{b},\ то\ должно\ \]

\[выполнять\ условие:\]

\[\overrightarrow{a} \bullet \overrightarrow{b} = 0.\]

\[\textbf{а)}\ \overrightarrow{a}\left\{ 4;5 \right\};\ \overrightarrow{b}\left\{ x; - 6 \right\}:\]

\[\overrightarrow{a} \bullet \overrightarrow{b} = 4x + 5 \bullet ( - 6) = 0\]

\[4x - 30 = 0\]

\[x = 7,5.\]

\[\textbf{б)}\ \overrightarrow{a}\left\{ x; - 1 \right\};\ \overrightarrow{b}\left\{ 3;2 \right\}:\]

\[\overrightarrow{a} \bullet \overrightarrow{b} = x \bullet 3 + ( - 1) \bullet 2 = 0\]

\[3x - 2 = 0\]

\[x = \frac{2}{3}.\]

\[\textbf{в)}\ \overrightarrow{a}\left\{ 0; - 3 \right\};\ \overrightarrow{b}\left\{ 5;x \right\}:\]

\[\overrightarrow{a} \bullet \overrightarrow{b} = 0 \bullet 5 + ( - 3)x = 0\]

\[- 3x = 0\]

\[x = 0.\]

\[\boxed{\mathbf{1047.еуроки - ответы\ на\ пятёрку}}\]

\[\mathbf{Дан}о:\]

\[A( - 2;4);B( - 5; - 3);\]

\[C( - 7; - 2);D(1;5);\]

\[(x + 5)^{2} + (y - 1)^{2} = 16 \Longrightarrow\]

\[\Longrightarrow O( - 5;1);\ R = 4.\]

\[1)\ A( - 2;4):\]

\[( - 2 + 5)^{2} + (4 - 1)^{2} \neq 16;\]

\[9 + 9 \neq 16;\]

\[18 \neq 16 \Longrightarrow \ A( - 2;4) -\]

\[вне\ круга.\]

\[2)\ B( - 5; - 3):\]

\[( - 5 + 5)^{2} + ( - 3 - 1)^{2} = 16;\]

\[16 = 16 \Longrightarrow B( - 5; - 3) -\]

\[на\ окружности.\]

\[3)\ C( - 7; - 2):\]

\[( - 7 + 5)^{2} + ( - 2 - 1)^{2} \neq 16;\ \]

\[4 + 9 \neq 16;\]

\[13 \neq 16 \Longrightarrow C( - 7; - 2) -\]

\[внутри\ круга.\]

\[4)\ D(1;5):\]

\[(1 + 5)^{2} + (5 - 1)^{2} \neq 16;\]

\[36 + 16 \neq 16\]

\[52 \neq 16 \Longrightarrow \ D(1;5) -\]

\[вне\ круга.\]

\[Ответ:а)\ C;б)\ B;в)\ A\ и\ \text{D.}\]