Cho đường thẳng d có phương trì...

D

- Hướng dẫn giải

Chọn D

TH1:

Ta có ${Đ}_O:M\left(x;y\right) \to M^{\text{'}}(x^{\text{'}};y^{\text{'}}).$  Khi đó:$\left\{\begin{array}{l}x\text{'}=-x\\ y\text{'}=-y\end{array}\right.\iff \left\{\begin{array}{l}x=-x^{\text{'}}\\ y=-y^{\text{'}}\end{array}\right.$

Từ $x+y-2=0\iff -x^{\text{'}}-y^{\text{'}}-2=0$

Vậy có ảnh $d_1:x+y+2=0$ .

Tiếp tục qua phép tịnh tiến $\overrightarrow{v}=\left(\mathrm{3,2}\right)$  có $T_{\overrightarrow{v}}:N\left(x;y\right)\to N^{\text{'}}\left(x^{\text{'}};y^{\text{'}}\right)$  khi đó  $\left\{\begin{array}{l}{x}^{\text{'}}=x+3\\ y^{\text{'}}=y+2\end{array}\right.\iff \left\{\begin{array}{l}x=3-x^{\text{'}}\\ y=2-y^{\text{'}}\end{array}\right.$.

$x+y+2=0$    $\iff \left(3-x^{\text{'}}\right)+\left(2-y^{\text{'}}\right)+2=0$$\iff 7-x^{\text{'}}-y^{\text{'}}=0$

Vậy ảnh là $d^{\text{'}}:x+y-7=0$ .

TH2:

Ta có qua phép tịnh tiến $\overrightarrow{v}=\left(\mathrm{3,2}\right)$  có $T_{\overrightarrow{v}}:N\left(x;y\right)\to N^{\text{'}}\left(x^{\text{'}};y^{\text{'}}\right)$  khi đó $\left\{\begin{array}{l}{x}^{\text{'}}=x+3\\ y^{\text{'}}=y+2\end{array}\right.\iff \left\{\begin{array}{l}x=3-x^{\text{'}}\\ y=2-y^{\text{'}}\end{array}\right.$  . Từ      $x+y-2=0$ $\iff \left(3-x^{\text{'}}\right)+\left(2-y^{\text{'}}\right)-2=0$$\iff 3-x^{\text{'}}-y^{\text{'}}=0$

Vậy có ảnh $d_1:x+y-3=0$ .

Tiếp tục ${Đ}_O:M\left(x;y\right) \to M^{\text{'}}(x^{\text{'}};y^{\text{'}}).$  Khi đó:$\left\{\begin{array}{l}x\text{'}=-x\\ y\text{'}=-y\end{array}\right.\iff \left\{\begin{array}{l}x=-x^{\text{'}}\\ y=-y^{\text{'}}\end{array}\right.$

Từ$x+y-3=0\iff -x^{\text{'}}-y^{\text{'}}-3=0$

Vậy ảnh là$d^{\text{'}}:x+y+3=0$ .