ГДЗ по геометрии 8 класс Атанасян Задание 948

\[\boxed{\mathbf{948.ОК\ ГДЗ - домашка\ на}\ 5}\]

\[\mathbf{Дано:}\]

\[\textbf{а)}\ A( - 3;5);\]

\[B(6;4);\]

\[C \in OY;\ \]

\[AC = CB.\]

\[\textbf{б)}\ C(4; - 3);\]

\[D(8;1);\]

\[E \in OY;\]

\[CE = ED.\]

\[\mathbf{Найти:}\]

\[\textbf{а)}\ точки\ C;\]

\[\textbf{б)}\ точки\ E.\]

\[\mathbf{Решение.}\]

\[\textbf{а)}\ 1)\ x_{C} = 0\ (так\ как\ C \in OY).\]

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

\[= \sqrt{9 + (y - 5)^{2}};\]

\[CB = \sqrt{(6 - 0)^{2} + (y - 4)^{2}} =\]

\[= \sqrt{36 + (y - 4)^{2}}.\]

\[3)\ AC = CB\ (по\ условию):\]

\[\sqrt{9 + (y - 5)^{2}} = \sqrt{36 + (y - 4)^{2}}\ \]

\[9 + (y - 5)^{2} = 36 + (y - 4)^{2}\]

\[(y - 5)^{2} - (y - 4)^{2} = 36 - 9\]

\[(y - 5 - y + 4)(y - 5 + y - 4) =\]

\[= 27\]

\[- 1(2y - 9) = 27\]

\[2y - 9 = - 27\]

\[2y = - 18\]

\[y = - 9.\]

\[Следовательно:\]

\[C(0; - 9).\]

\[\textbf{б)}\ 1)\ x_{E} = 0\ (так\ как\ E \in OY).\]

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

\[= \sqrt{16 + (y + 3)^{2}};\]

\[ED = \sqrt{(8 - 0)^{2} + (1 - y)^{2}} =\]

\[= \sqrt{64 + (1 - y)^{2}}.\]

\[3)\ CE = ED\ (по\ условию):\]

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

\[= \sqrt{64 + (1 - y)^{2}}\ \]

\[16 + (y + 3)^{2} = 64 + (1 - y)^{2}\]

\[(y + 3)^{2} - (1 - y)^{2} = 64 - 16\]

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

\[= 48\]

\[4(2y + 2) = 48\]

\[8(y + 1) = 48\]

\[y + 1 = 6\]

\[y = 5.\]

\[Следовательно:\]

\[E(0;5).\]

\[\mathbf{Ответ:}\mathbf{\ }а)\ C(0; - 9);\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ E(0;5).\]