Решебник по геометрии 11 класс. Атанасян ФГОС 662

\[\boxed{\mathbf{662.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]

\[Формула\ для\ расчета\ \]

\[координат\ середины\ отрезка:\]

\[M\left( \frac{x_{a} + x_{b}}{2};\ \frac{y_{a} + y_{b}}{2};\ \frac{z_{a} + z_{b}}{2} \right).\]

\[\frac{x_{a} + x_{b}}{2} = x_{m}\]

\[x_{a} = 2x_{m} - x_{b};\]

\[x_{b} = 2x_{m} - x_{a}.\]

\[AM = MB;\ \ M(x;0;0).\]

\[\textbf{а)}\ A( - 3;m;5);\ B(2; - 2;n).\]

\[m = 2 \cdot 0 - ( - 2) = 2;\]

\[n = 2 \cdot 0 - 5 = - 5;\]

\[m = 2;\ \ n = - 5.\]

\[\textbf{б)}\ A(1;0,5; - 4);\ \ B(1;m;2n).\]

\[m = 2 \cdot 0 - 0,5 = - 0,5;\]

\[2n = 2 \cdot 0 - ( - 4) = 4;\]

\[m = - 0,5;\ \ n = 2.\]

\[\textbf{в)}\ A(0;m;n + 1);\ \ \]

\[B(1;n; - m + 1).\]

\[m = 2 \cdot 0 - n = - n;\]

\[n + 1 = 2 \cdot 0 - ( - m + 1) =\]

\[= m - 1 = - n - 1;\]

\[n = - 1;\ \ m = 1.\]

\[\textbf{г)}\ A(7;2m + n; - n);\ \ \]

\[B( - 5; - 3;m - 3).\]

\[2m + n = 2 \cdot 0 - ( - 3) = 3;\]

\[- n = 2 \cdot 0 - (m - 3) = - m + 3;\]

\[m = 2;\ \ n = - 1.\]