To reflect one point about a horizontal line, we want the image and the preimage to be equidistant to that line.
Then, to reflect for example point (8,8) about the line y = -1, we need to find a point with the same x-coordinate and whose distance to line y = -1 is the same.
The distance from point 8,8 to line y = -1 is 8 - (-1) = 8 + 1 = 9.
Then, a point that is 9 units down the line y = -1 will be (8,-10) since -1 - (-10) = -1 + 10 = 9.
We found then that to reflect about a line y = a, we need to add twice a to the negative y coordinate:
[tex](x,y)\rightarrow(x,-y+2a)[/tex]Then, for reflection about y = -1:
[tex]\begin{gathered} A(8,8)\rightarrow A^{\prime}(8,-8+2(-1))=A^{\prime}(8,-8-2) \\ A(8,8)\rightarrow A(8,-10) \end{gathered}[/tex][tex]\begin{gathered} B(10,6)\rightarrow B^{\prime}(10,-6+2(-1))=B^{\prime}(8,-6-2) \\ B(10,6)\rightarrow B^{\prime}(10,-8) \end{gathered}[/tex][tex]\begin{gathered} C(2,2)\rightarrow C^{\prime}(2,-2+2(-1))=C^{\prime}(2,-2-2) \\ C(2,2)\rightarrow C^{\prime}(2,-4) \end{gathered}[/tex]Now we need to use the same equation:
[tex](x,y)\rightarrow(x,-y+2a)[/tex]To transform the new images, which are:
A'(8,-10)
B'(10,-8)
C'(2,-4)
Applying the same for them:
[tex]\begin{gathered} A^{\prime}(8,-10)\rightarrow A´´(8,-(-10)+2(-7))=A´´(8,10-14) \\ A^{\prime}(8,-10)\rightarrow A´´(8,-4) \end{gathered}[/tex][tex]\begin{gathered} B^{\prime}(10,-8)\rightarrow B´´(10,-(-8)+2(-7))=B´´(10,8-14) \\ B^{\prime}(10,-8)\rightarrow B´´(10,-6) \end{gathered}[/tex][tex]\begin{gathered} C^{\prime}(2,-4)\rightarrow C´´(2,-(-4)+2(-7))=C´´(2,4-14) \\ C^{\prime}(2,-4)\rightarrow C´´(2,-10) \end{gathered}[/tex]Now we have all the blanks:
For the reflection about y = -1:
A'(8, -10)
B'(10, -8)
C'(2, -4)
For the reflection about y = -7:
A'(8, -4)
B'(10, -6)
C'(2, -10)
