\(\eqalign{
& Lay\,\,A\left( { - 2;0} \right);\,\,B\left( {1; - 2} \right) \cr
& \left\{ \matrix{
x' = \left( {x - a} \right)\cos \alpha - \left( {y - b} \right)\sin \alpha + a \hfill \cr
y' = \left( {y - a} \right)\sin \alpha + \left( {y - b} \right)\cos \alpha + b \hfill \cr} \right. \cr
& A' = {Q_{\left( {I;{{45}^0}} \right)}}\left( A \right) \cr
& \Leftrightarrow \left\{ \matrix{
x' = \left( { - 2 - 1} \right)\cos 45 - \left( {0 - 2} \right)\sin 45 + 1 \hfill \cr
y' = \left( { - 2 - 1} \right)\sin 45 + \left( {0 - 2} \right)\cos 45 + 2 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x' = - 3{{\sqrt 2 } \over 2} + 2{{\sqrt 2 } \over 2} + 1 = - {{\sqrt 2 } \over 2} + 1 \hfill \cr
y' = - 3{{\sqrt 2 } \over 2} - 2{{\sqrt 2 } \over 2} + 2 = - {{5\sqrt 2 } \over 2} + 2 \hfill \cr} \right. \cr
& \Rightarrow A'\left( { - {{\sqrt 2 } \over 2} + 1; - {{5\sqrt 2 } \over 2} + 2} \right) \cr
& B' = {Q_{\left( {I;45} \right)}}\left( B \right) \cr
& \Rightarrow \left\{ \matrix{
x' = \left( {1 - 1} \right)\cos 45 - \left( { - 2 - 2} \right)\sin 45 + 1 \hfill \cr
y' = \left( {1 - 1} \right)\sin 45 + \left( { - 2 - 2} \right)\sin 45 + 2 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x' = 4{{\sqrt 2 } \over 2} + 1 = 2\sqrt 2 + 1 \hfill \cr
y' = - 4{{\sqrt 2 } \over 2} + 2 = - 2\sqrt 2 + 2 \hfill \cr} \right. \cr
& \Rightarrow B'\left( {2\sqrt 2 + 1; - 2\sqrt 2 + 2} \right) \cr} \)
Sau đó viết phương trình đường thẳng đi qua 2 điểm A' và B'
