A ___ C ___ D ___ B
Значит, С - середина AD
![x_{C}=\frac{x_{A}+x_{D}}{2}\Rightarrow x_{A}=2x_{C}-x_{D}=2\cdot 2-5=-1 x_{C}=\frac{x_{A}+x_{D}}{2}\Rightarrow x_{A}=2x_{C}-x_{D}=2\cdot 2-5=-1](https://tex.z-dn.net/?f=x_%7BC%7D%3D%5Cfrac%7Bx_%7BA%7D%2Bx_%7BD%7D%7D%7B2%7D%5CRightarrow%20x_%7BA%7D%3D2x_%7BC%7D-x_%7BD%7D%3D2%5Ccdot%202-5%3D-1)
Аналогично
![y_{A}=2y_{C}-y_{D}=2\cdot 0-(-2)=2\\ \\z_{A}=2z_{C}-z_{D}=2\cdot 2-0=4 y_{A}=2y_{C}-y_{D}=2\cdot 0-(-2)=2\\ \\z_{A}=2z_{C}-z_{D}=2\cdot 2-0=4](https://tex.z-dn.net/?f=y_%7BA%7D%3D2y_%7BC%7D-y_%7BD%7D%3D2%5Ccdot%200-%28-2%29%3D2%5C%5C%20%5C%5Cz_%7BA%7D%3D2z_%7BC%7D-z_%7BD%7D%3D2%5Ccdot%202-0%3D4)
D - середина CB
x_{D}=\frac{x_{C}+x_{B}}{2}\Rightarrow x_{B}=2x_{D}-x_{C}=2\cdot 5-2=8[/tex]
y_{B}=2y_{D}-y_{C}=2\cdot (-2)-0=-4\\ \\z_{B}=2z_{D}-z_{C}=2\cdot 0-2=-2[/tex]
О т в е т. A(-1;2;4); B(8;4;-2)