\[\boxed{\mathbf{936.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[AM = MB \Longrightarrow M - середина\ \text{AB.}\]
\[A\] | \[(2; - 3)\] | \[( - 10; - 11)\] | \[(0;1)\] | \[(0;0)\] | \[(c,d)\] | \[(3;5)\] | \[(3t + 5;7)\] | \[(1;3)\] |
---|---|---|---|---|---|---|---|---|
\[B\] | \[( - 3;1)\] | \[(4;7)\] | \[(6; - 11)\] | \[( - 3;7)\] | \[(2a - c;2b - d)\] | \[(3;8)\] | \[(t + 7; - 7)\] | \[( - 1; - 3)\] |
\[M\] | \[( - 0,5; - 1)\] | \[( - 3; - 2)\] | \[(3; - 5)\] | \[( - 1,5;3,5)\] | \[(a;b)\] | \[(3;6,5)\] | \[(2t + 6;0)\] | \[(0;0)\] |
\[\boxed{\mathbf{936.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\]
\[ABCD - параллелограмм;\]
\[\text{AC\ }и\ BD - диагонали;\]
\[AC \cap BD = O.\]
\[Равны\ ли\ векторы:\]
\[\textbf{а)}\ \overrightarrow{\text{AB}}\ и\ \overrightarrow{\text{DC}}:\]
\[\overrightarrow{\text{AB}} = \overrightarrow{\text{DC}}.\]
\[\textbf{б)}\ \overrightarrow{\text{BC}}\ и\ \overrightarrow{\text{DA}}:\]
\[\overrightarrow{\text{BC}} \neq \overrightarrow{\text{DA}}.\]
\[\textbf{в)}\ \overrightarrow{\text{AO}}\ и\ \overrightarrow{\text{OC}}:\]
\[\overrightarrow{\text{AO}} = \overrightarrow{\text{OC}}.\]
\[\textbf{г)}\ \overrightarrow{\text{AC}}\ и\ \overrightarrow{\text{BD}}:\]
\[\overrightarrow{\text{AC}}\ и\ \overrightarrow{\text{BD}} - не\ коллинеарны;\]
\[\overrightarrow{\text{AC}} \neq \overrightarrow{\text{BD}}.\]
\[Ответ:а)\ да;б)\ нет;в)\ да;\]
\[\textbf{г)}\ нет.\]