When a shape is reflected, it must be reflected over a line
The reflection rule is: [tex]R_{x = 2}(\triangle D EF) = \triangle D'E'F[/tex]
The coordinates of the preimage are:
D(3,6), E(-4,-3), F(6,1)
The coordinates of the image are:
D'(1,6), E'(8,-3), F'(-2,1)
From the given coordinates, we have the following observations
This is shown as follows:
D(3,6) and D'(1,6)
The y-coordinate is:
[tex]y = 6[/tex]
The average of the x-coordinates
[tex]x = \frac{3 + 1}{2}[/tex]
[tex]x = 2[/tex]
E(-4,-3) and E'(8,-3)
The y-coordinate is:
[tex]y = -3[/tex]
The average of the x-coordinates
[tex]x = \frac{-4 + 8}{2}[/tex]
[tex]x = 2[/tex]
F(6,1) and F'(-2,1)
The y-coordinate is:
[tex]y =1[/tex]
The average of the x-coordinates
[tex]x = \frac{6 -2}{2}[/tex]
[tex]x = 2[/tex]
Since the y-coordinates remains unchanged and the average of the x-coordinates is 2.
Then, we can conclude that the triangle was reflected over line [tex]x = 2[/tex]
Hence, the reflection rule is:
[tex]R_{x = 2}(\triangle D EF) = \triangle D'E'F[/tex]
Read more about reflections at:
https://brainly.com/question/17983440
