Решебник по геометрии 11 класс. Атанасян ФГОС 672

\[\boxed{\mathbf{672.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Дано:\]

\[⊿ABC;\]

\[A(1;0;k);\ \ B( - 1;2;3);\ \ \]

\[C(0;0;1).\]

\[Найдем\ длины\ сторон\ \]

\[треугольника\ по\ формуле:\]

\[\left| \text{AB} \right| = \left| \overrightarrow{\text{AB}} \right| =\]

\[= \sqrt{( - 1 - 1)^{2} + (2 - 0)^{2} + (3 - k)^{2}} =\]

\[= \sqrt{17 - 6k + k^{2}};\]

\[\left| \text{AC} \right| = \left| \overrightarrow{\text{AC}} \right| =\]

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

\[= \sqrt{2 - 2k + k^{2}};\]

\[\left| \text{BC} \right| = \left| \overrightarrow{\text{BC}} \right| =\]

\[= \sqrt{(0 - 1)^{2} + (0 - 2)^{2} + (1 - 3)^{2}} =\]

\[= \sqrt{9} = 3.\]

\[⊿ABC - равнобедренный,\ \]

\[стороны\ попарно\ равны.\]

\[Имеем\ три\ возможных\ \]

\[варианта.\]

\[1)\ AB = AC:\]

\[17 - 6k + k^{2} = 2 - 2k + k^{2}\]

\[15 = 4k\]

\[k = 3,75.\]

\[2)\ AB = BC:\]

\[17 - 6k + k^{2} = 9\]

\[k^{2} - 6k + 8 = 0\]

\[D_{1} = 9 - 8 = 1\]

\[k_{1} = 3 + 1 = 4;\ \ k_{2} = 3 - 1 = 2.\]

\[3)\ BC = AC:\]

\[9 = 2 - 2k + k^{2}\]

\[k^{2} - 2k - 7 = 0\]

\[D_{1} = 1 + 7 = 8\]

\[k_{1} = 1 + \sqrt{8} = 1 + 2\sqrt{2};\ \ \ \]

\[k_{2} = 1 - \sqrt{8} = 1 - 2\sqrt{2}.\]

\[Ответ:\]

\[k = \left\{ 1 - 2\sqrt{2};\ \ 2;3,75;1 + 2\sqrt{2};\ \ 4 \right\}.\]