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МатематикаРешебник по геометрии 10 класс Атанасян ФГОС 674
674
\[\boxed{\mathbf{674.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[A( - 2;3;5);\ \ B(3;2; - 3).\]
\[\textbf{а)}\ K(x;0;0) - точка\ по\ оси\ \text{Ox},\ \]
\[равноудаленная\ от\ точек\ \]
\[\text{A\ }и\ B:\]
\[KA = KB.\]
\[\left| \overrightarrow{\text{KA}} \right| = \sqrt{( - 2 - x)^{2} + 3^{2} + 5^{2}} =\]
\[= \sqrt{4 + 4x + x^{2} + 9 + 25} =\]
\[= \sqrt{x^{2} + 4x + 38};\]
\[\left| \overrightarrow{\text{KB}} \right| =\]
\[= \sqrt{(3 - x)^{2} + 2^{2} + ( - 3)^{2}} =\]
\[= \sqrt{9 - 6x + x^{2} + 4 + 9} =\]
\[= \sqrt{x^{2} - 6x + 22};\]
\[x^{2} + 4x + 38 = x^{2} - 6x + 22\]
\[10x = - 16\]
\[x = - 1,6.\]
\[K( - 1,6;0;0).\]
\[\textbf{б)}D(0;y;0) - точка\ по\ оси\ \text{Oy},\ \]
\[равноудаленная\ от\ точек\ \text{A\ }и\ B:\]
\[AD = DB.\]
\[\left| \overrightarrow{\text{AD}} \right| =\]
\[= \sqrt{( - 2 - 0)^{2} + (3 - y)^{2} + 5^{2}} =\]
\[= \sqrt{4 + 9 - 6y + y^{2} + 25} =\]
\[= \sqrt{y^{2} - 6y + 38};\ \]
\[\left| \overrightarrow{\text{DB}} \right| =\]
\[= \sqrt{3^{2} + (2 - y)^{2} + ( - 3)^{2}} =\]
\[= \sqrt{9 + 4 - 4y + y^{2} + 9} =\]
\[= \sqrt{y^{2} - 4y + 22};\]
\[y^{2} - 6y + 38 = y^{2} - 4y + 22\]
\[- 2y = - 16\]
\[y = 8.\]
\[D(0;8;0).\]
\[\textbf{в)}\ C(0;0;z) - точка\ по\ оси\ \text{Oz},\ \]
\[равноудаленная\ от\ точек\ \text{A\ }и\ B:\]
\[AC = CB.\]
\[\left| \overrightarrow{\text{AC}} \right| =\]
\[= \sqrt{( - 2)^{2} + 3^{2} + (5 - z)^{2}} =\]
\[= \sqrt{13 + 25 - 10z + z^{2}} =\]
\[= \sqrt{z^{2} - 10z + 38};\]
\[\left| \overrightarrow{\text{CB}} \right| = \sqrt{3^{2} + 2^{2} + ( - 3 - z)^{2}} =\]
\[= \sqrt{13 + 9 + 6z + z^{2}} =\]
\[= \sqrt{z^{2} + 6z + 22};\]
\[z^{2} - 10z + 38 = z^{2} + 6z + 22\]
\[- 16z = - 16\]
\[z = 1.\]
\[C(0;0;1).\]
