Given a triangle made with the vertices A(1,1), B(3,1), and C(3,2). Apply the following rules rule (x,y)-->(x, -y+6) to triangle ABC. Give me the coordinates of A', B', and c'

Answer :

Answer:

A'(1, 5)

B'(3, 5)

C'(3, 4)

Explanation:

Given the vertices of triangle ABC to be A(1,1), B(3,1), and C(3,2), let's go ahead and apply the rule (x,y)-->(x, -y+6) to have triangle A'B'C';

[tex]\begin{gathered} A(1,1)\rightarrow A^{\prime}(1,-1+6) \\ \therefore A(1,1)\rightarrow A^{\prime}(1,5) \\ \\ \end{gathered}[/tex][tex]\begin{gathered} B(3,1)\rightarrow B^{\prime}(3,-1+6) \\ \therefore B(3,1)\rightarrow B^{\prime}(3,5) \end{gathered}[/tex][tex]\begin{gathered} C(3,2)\rightarrow C^{\prime}(3,-2+6) \\ \therefore C(3,2)\rightarrow C^{\prime}(3,4) \end{gathered}[/tex]