Question

Reflect the given preimage over =−1y=−1 followed by =−7y=−7. Find the new coordinates. What one transformation is this double reflection the same as?

Answer 1

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:

`(x,y)\rightarrow(x,-y+2a)`

Then, for reflection about y = -1:

`\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}`
`\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}`
`\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}`

Now we need to use the same equation:

`(x,y)\rightarrow(x,-y+2a)`

To transform the new images, which are:

A'(8,-10)

B'(10,-8)

C'(2,-4)

Applying the same for them:

`\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}`
`\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}`

`\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}`

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)