\[\boxed{\mathbf{949.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Дано:}\]
\[\textbf{а)}\ A(1;2);\]
\[B( - 3;4);\]
\[E \in OX;\ \]
\[AE = CB.\]
\[\textbf{б)}\ C(1;1);\]
\[D(3;5);\]
\[F \in OX;\ \]
\[CF = FD.\]
\[\mathbf{Найти:}\]
\[координаты\]
\[\textbf{а)}\ точки\ E;\]
\[\textbf{б)}\ точки\ F.\]
\[\mathbf{Решение.}\]
\[\textbf{а)}\ 1)\ y_{E} = 0\ (так\ как\ E \in OX).\]
\[2)\ AE = \sqrt{(x - 1)^{2} + (0 - 2)^{2}} =\]
\[= \sqrt{(x - 1)^{2} + 4};\]
\[EB = \sqrt{( - 3 - x)^{2} + (4 - 0)^{2}} =\]
\[= \sqrt{( - 3 - x)^{2} + 16}.\]
\[3)\ AE = EB\ (по\ условию):\]
\[\sqrt{(x - 1)^{2} + 4} =\]
\[= \sqrt{( - 3 - x)^{2} + 16}\]
\[(x - 1)^{2} + 4 = ( - 3 - x)^{2} + 16\]
\[(x - 1)^{2} - ( - 3 - x)^{2} = 16 - 4\]
\[(x - 1 + 3 + x)(x - 1 - 3 - x) =\]
\[= 12\]
\[- 4(2x + 2) = 12\]
\[2x + 2 = - 3\]
\[2x = - 5\]
\[x = - 2,5.\]
\[Следовательно:\]
\[E( - 2,5;0)\]
\[\textbf{б)}\ 1)\ y_{F} = 0\ (так\ как\ F \in OX).\]
\[2)\ CF = \sqrt{(x - 1)^{2} + (0 - 1)^{2}} =\]
\[= \sqrt{(x - 1)^{2} + 1};\]
\[FD = \sqrt{(3 - x)^{2} + (5 - 0)^{2}} =\]
\[= \sqrt{(3 - x)^{2} + 25}.\]
\[3)\ CF = FD\ (по\ условию):\]
\[\sqrt{(x - 1)^{2} + 1} = \sqrt{(3 - x)^{2} + 25}\ \]
\[(x - 1)^{2} + 1 = (3 - x)^{2} + 25\]
\[(x - 1)^{2} - (3 - x)^{2} = 25 - 1\]
\[(x - 1 - 3 + x)(x - 1 + 3 - x) =\]
\[= 24\]
\[- 2(2x - 4) = 24\]
\[2x - 4 = 12\]
\[2x = 16\]
\[x = 8.\]
\[Следовательно:\]
\[F(8;0)\]
\[\mathbf{Ответ:}\mathbf{\ }а)\ E( - 2,5;0);\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\text{\ F}(8;0).\]
\[\boxed{\mathbf{949.еуроки - ответы\ на\ пятёрку}}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\]
\[A,B,C\ и\ D - произвольные\ \]
\[точки.\]
\[Доказать:\ \]
\[\overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} + \overrightarrow{\text{CD}} + \overrightarrow{\text{DA}} = \overrightarrow{0}.\]
\[Доказательство.\]
\[1)\ \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} + \overrightarrow{\text{CD}} + \overrightarrow{\text{DA}} =\]
\[= \overrightarrow{\text{AA}}\ (по\ правилу\ многоугольника).\]
\[3)\ Следовательно:\]
\[\overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} + \overrightarrow{\text{CD}} + \overrightarrow{\text{DA}} = \overrightarrow{0}.\]
\[Что\ и\ требовалось\ доказать.\ \]