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МатематикаРешебник по геометрии 10 класс Атанасян ФГОС 675
675
\[\boxed{\mathbf{675.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[A( - 1;2;3);\ \ B( - 2;1;2);\ \ \]
\[C(0; - 1;1).\]
\[\textbf{а)}\ Oxy.\]
\[K(x;y;z) - равноудалена\ от\ \]
\[точек\ A;B;C:\]
\[\left| \overrightarrow{\text{AK}} \right| =\]
\[= \sqrt{( - 1 - x)^{2} + (2 - y)^{2} + 3^{2}}\]
\[\left| \overrightarrow{\text{BK}} \right| =\]
\[= \sqrt{( - 2 - x)^{2} + (1 - y)^{2} + 2^{2}}\]
\[\left| \overrightarrow{\text{CK}} \right| = \sqrt{x^{2} + ( - 1 - y)^{2} + 1^{2}}\]
\[AK = BK:\]
\[1 + 2x + x^{2} + 4 - 4y + y^{2} + 9 =\]
\[= 4 + 4x + x^{2} + 1 - 2y + y^{2} + 4\]
\[2x - 4y + 14 = 4x - 2y + 9\]
\[2x + 2y = 5\]
\[AK = CK:\]
\[2x - 4y + 14 =\]
\[= x^{2} + 1 + 2y + y^{2} + 1\]
\[2x - 4y + 14 = 2y + 2\]
\[2x - 6y = - 12\]
\[\left\{ \begin{matrix} 2x + 2y = 5\ \ \ \ \ \\ 2x - 6y = - 12 \\ \end{matrix} \right.\ ( - )\]
\[8y = 17\]
\[y = \frac{17}{8}.\]
\[2x + 2 \cdot \frac{17}{8} = 5\]
\[2x = 5 - \frac{17}{4}\]
\[2x = \frac{3}{4}\]
\[x = \frac{3}{8}.\]
\[K\left( \frac{3}{8};\ \frac{17}{8};0 \right).\]
\[\textbf{б)}\ Oyz.\]
\[P(0;y;z) - равноудалена\ от\ \]
\[точек\ A;B;C:\]
\[AP = BP;\ \ AP = CP;\ \ BP = CP.\]
\[\left| \overrightarrow{\text{AP}} \right| =\]
\[= \sqrt{( - 1)^{2} + (2 - y)^{2} + (3 - z)^{2}};\]
\[\left| \overrightarrow{\text{BP}} \right| =\]
\[= \sqrt{( - 2)^{2} + (1 - y)^{2} + (2 - z)^{2}};\]
\[\left| \overrightarrow{\text{CP}} \right| =\]
\[= \sqrt{0 + ( - 1 - y)^{2} + (1 - z)^{2}.}\]
\[AP = BP:\]
\[1 + 4 - 4y + y^{2} + 9 - 6z + z^{2} =\]
\[= 4 + 1 - 2y + y^{2} + 4 - 4z + z^{2}\]
\[- 4y - 6z + 14 = - 2y - 4z + 9\]
\[2y + 2z = 5.\]
\[BP = CP:\]
\[- 4y - 6z + 14 =\]
\[= 1 + 2y + y^{2} + 1 - 2z + z^{2}\]
\[- 4y - 6z + 14 = 2y - 2z + 2\]
\[6y + 4z = 12\ \ |\ :2\]
\[3y + 2z = 6.\]
\[\left\{ \begin{matrix} 2y + 2z = 5 \\ 3y + 2z = 6 \\ \end{matrix} \right.\ ( - )\]
\[- y = - 1\]
\[y = 1.\]
\[2z = 5 - 2y = 5 - 2 \cdot 1 = 3\]
\[z = 1,5.\]
\[P(0;1;1,5).\]
\[\textbf{в)}\ Ozx.\]
\[R(x;0;z) - равноудалена\ от\ \ \]
\[точек\ A;B;C:\]
\[AR = BR;\ \ AR = CR.\]
\[\left| \overrightarrow{\text{AR}} \right| =\]
\[= \sqrt{( - 1 - x)^{2} + 2^{2} + (3 - z)^{2}};\]
\[\left| \overrightarrow{\text{BR}} \right| =\]
\[= \sqrt{( - 2 - x)^{2} + 1^{2} + (2 - z)^{2}};\]
\[\left| \overrightarrow{\text{CR}} \right| =\]
\[= \sqrt{x^{2} + ( - 1)^{2} + (1 - z)^{2}}.\]
\[AR = BR:\]
\[1 + 2x + x^{2} + 4 + 9 - 6z + z^{2} =\]
\[= 4 + 4x + x^{2} + 1 + 4 - 4z + z^{2}\]
\[2x - 6z + 14 = 4x - 4z + 9\]
\[2x + 2z = 5.\]
\[AR = CR:\]
\[2x - 6z + 14 =\]
\[= x^{2} + 1 + 1 - 2z + z^{2}\]
\[2x - 4z = - 12.\]
\[\left\{ \begin{matrix} 2x + 2z = 5\ \ \ \ \ \\ 2x - 4z = - 12 \\ \end{matrix} \right.\ ( - )\]
\[6z = 17\]
\[z = \frac{17}{6}.\]
\[2x + 2 \cdot \frac{17}{6} = 5\]
\[2x = 5 - \frac{17}{3} = - \frac{2}{3}\]
\[x = - \frac{1}{3}.\]
\[R\left( - \frac{1}{3};0;\frac{17}{6} \right).\]
