Question

answer all parts of the question on the graph to the right. plot the pre - image and label it appropriately. make sure to show the transformations using different colors and labeling them appropriately. show the line of reflection when reflecting.
. plot the pre - image. pentagon efghi: e(-5,1), f(-3,3), g(-1,3), h(1,1), i(-2,0).
. then, reflect the pentagon across the line y = 3 creating image efghi.
. then, rotate efghi 90° counter clockwise about the origin to create image efghi

Explanation:

Step1: Reflection across $y = 3$

The formula for reflecting a point $(x,y)$ across the line $y = k$ is $(x,2k - y)$. Here $k = 3$.
For point E(-5,1):
$x=-5,y = 1$, new $y=2\times3 - 1=5$, so E' is (-5,5).
For point F(-3,3):
$x=-3,y = 3$, new $y=2\times3 - 3=3$, so F' is (-3,3).
For point G(-1,3):
$x=-1,y = 3$, new $y=2\times3 - 3=3$, so G' is (-1,3).
For point H(1,1):
$x = 1,y=1$, new $y=2\times3 - 1=5$, so H' is (1,5).
For point I(-2,0):
$x=-2,y = 0$, new $y=2\times3 - 0=6$, so I' is (-2,6).

Step2: Rotation 90° counter - clockwise about the origin

The formula for rotating a point $(x,y)$ 90° counter - clockwise about the origin is $(-y,x)$.
For point E'(-5,5):
$x=-5,y = 5$, new point E'' is (-5,-5).
For point F'(-3,3):
$x=-3,y = 3$, new point F'' is (-3,-3).
For point G'(-1,3):
$x=-1,y = 3$, new point G'' is (-3,-1).
For point H'(1,5):
$x = 1,y=5$, new point H'' is (-5,1).
For point I'(-2,6):
$x=-2,y = 6$, new point I'' is (-6,-2).

Answer:

Pre - image points: E(-5,1), F(-3,3), G(-1,3), H(1,1), I(-2,0)
Image after reflection E'(-5,5), F'(-3,3), G'(-1,3), H'(1,5), I'(-2,6)
Image after rotation E''(-5,-5), F''(-3,-3), G''(-3,-1), H''(-5,1), I''(-6,-2)