ГДЗ по геометрии 9 класс Атанасян Задание 949

\[\boxed{\mathbf{949.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\]

\[\textbf{а)}\ A(1;2);\]

\[B( - 3;4);\]

\[E \in OX;\ \]

\[AE = CB.\]

\[\textbf{б)}\ C(1;1);\]

\[D(3;5);\]

\[F \in OX;\ \]

\[CF = FD.\]

\[\mathbf{Найти:}\]

\[координаты\]

\[\textbf{а)}\ точки\ E;\]

\[\textbf{б)}\ точки\ F.\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ 1)\ y_{E} = 0\ (так\ как\ E \in OX).\]

\[2)\ AE = \sqrt{(x - 1)^{2} + (0 - 2)^{2}} =\]

\[= \sqrt{(x - 1)^{2} + 4};\]

\[EB = \sqrt{( - 3 - x)^{2} + (4 - 0)^{2}} =\]

\[= \sqrt{( - 3 - x)^{2} + 16}.\]

\[3)\ AE = EB\ (по\ условию):\]

\[\sqrt{(x - 1)^{2} + 4} =\]

\[= \sqrt{( - 3 - x)^{2} + 16}\]

\[(x - 1)^{2} + 4 = ( - 3 - x)^{2} + 16\]

\[(x - 1)^{2} - ( - 3 - x)^{2} = 16 - 4\]

\[(x - 1 + 3 + x)(x - 1 - 3 - x) =\]

\[= 12\]

\[- 4(2x + 2) = 12\]

\[2x + 2 = - 3\]

\[2x = - 5\]

\[x = - 2,5.\]

\[Следовательно:\]

\[E( - 2,5;0)\]

\[\textbf{б)}\ 1)\ y_{F} = 0\ (так\ как\ F \in OX).\]

\[2)\ CF = \sqrt{(x - 1)^{2} + (0 - 1)^{2}} =\]

\[= \sqrt{(x - 1)^{2} + 1};\]

\[FD = \sqrt{(3 - x)^{2} + (5 - 0)^{2}} =\]

\[= \sqrt{(3 - x)^{2} + 25}.\]

\[3)\ CF = FD\ (по\ условию):\]

\[\sqrt{(x - 1)^{2} + 1} = \sqrt{(3 - x)^{2} + 25}\ \]

\[(x - 1)^{2} + 1 = (3 - x)^{2} + 25\]

\[(x - 1)^{2} - (3 - x)^{2} = 25 - 1\]

\[(x - 1 - 3 + x)(x - 1 + 3 - x) =\]

\[= 24\]

\[- 2(2x - 4) = 24\]

\[2x - 4 = 12\]

\[2x = 16\]

\[x = 8.\]

\[Следовательно:\]

\[F(8;0)\]

\[\mathbf{Ответ:}\mathbf{\ }а)\ E( - 2,5;0);\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\text{\ F}(8;0).\]