\(A\left(x_A;y_A\right)\Rightarrow A_1\left(2x_A-1;-y_A+3\right)\)
B(xB;yB) =>B1(2xB-1;-yB+3)
\(\overrightarrow{AB}=\left(x_B-x_A;y_B-y_A\right)\)
\(\overrightarrow{A_1B_1}=\left(2x_B-1-2x_A+1;-y_B+3+y_A-3\right)=\left(2x_B-2x_A;-y_B+y_A\right)\)
Vì \(\dfrac{x_B-x_A}{2\left(x_B-x_A\right)}< >\dfrac{y_B-y_A}{-\left(y_B-y_A\right)}\)
nên F ko là phép đồng dạng