Given
A(11,9)
B(11,3)
C(5,3)
D(5,9)
In the coordinate system (x,y), x determines the horizontal position, and y determines the vertical position. Having a translation of 9 units left, and 1 unit down, means that each coordinate system will be translated as (x-9 , y-1).
[tex]\begin{gathered} (x,y)\Longrightarrow(x-9,y-1) \\ \\ A(11,9)\Longrightarrow A^{\prime}(11-9,9-1)\Rightarrow A^{\prime}(2,8) \\ B(11,3)\operatorname{\Longrightarrow}B^{\prime}(11-9,3-1)\operatorname{\Rightarrow}B^{\prime}(2,2) \\ C(5,3)\operatorname{\Longrightarrow}C^{\prime}(5-9,3-1)\operatorname{\Rightarrow}C^{\prime}(-4,2) \\ D(5,9)\operatorname{\Longrightarrow}D^{\prime}(5-9,9-1)\operatorname{\Rightarrow}D^{\prime}(-4,8) \end{gathered}[/tex]
Therefore, the coordinates of the post image are A'(2,8), B'(2,2), C'(-4,2), D'(-4,8).
