Question

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'?

Answer 1

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.

`\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}`

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