The vertices of AABC are A (1, 1), B (5, 1), and C (3, 3). Draw the triangle and its image, AA'B'C', translated 5 units to the left. What are the coordinates of the vertices of AA'B'C'?

Respuesta :

The given transformation is a translation of 5 units to the left. So, we have to subtract each x-coordinate with the number of units given in the transformation.

[tex]\begin{gathered} A(1,1)\rightarrow A^{\prime}(1-5,1)\rightarrow A^{\prime}(-4,1) \\ B(5,1)\rightarrow B^{\prime}(5-5,1)\rightarrow B^{\prime}(0,1) \\ C(3,3)\rightarrow C^{\prime}(3-5,3)\rightarrow C^{\prime}(-2,3) \end{gathered}[/tex]

Hence, the new vertices are A'(-4,1), B'(0,1), and C'(-2,3).