Answer:
ABCD is reflected over the y-axis and translate (x + [-4] , y + [-4])
Explanation:
* Lets explain the reflection and the translation
- If point (x , y) reflected across the x-axis
∴ Its image is (x , -y)
- If point (x , y) reflected across the y-axis
∴ Its image is (-x , y)
- If the point (x , y) translated horizontally to the right by h units
∴ Its image is (x + h , y)
- If the point (x , y) translated horizontally to the left by h units
∴ Its image is (x - h , y)
- If the point (x , y) translated vertically up by k units
∴ Its image is (x , y + k)
- If the point (x , y) translated vertically down by k units
∴ Its image is (x , y - k)
* Now lets solve the problem
∵ The vertices of ABCD are:
A = (-5 , 2) , B = (-3 , 4) , C = (-2 , 4) , D = (-1 , 2)
- If ABCD reflected over the y-axis we will change the sign of the
x-coordinate of all the points
∴ The image of A = (5 , 2)
∴ The image of B = (3 , 4)
∴ The image of C = (2 , 4)
∴ The image of D = (1 , 2)
∴ The image of ABCD after reflection over the y-axis will be at
the first quadrant
- The figure EHGF is left and down the image of ABCD after
the reflection over the y-axis
∴ The image is translate to the left and down
∵ E = (1 , -2) and E is the image of A after reflection and translation
∵ The image of a after reflection is (5 , 2)
- That means 5 became 1 and 2 became -2
∵ 1 - 5 = -4
∴ The image of A after reflection translate 4 units to the left
∵ -2 - 2 = -4
∴ The image of A after reflection translate 4 units down
∴ ABCD is reflected over the y-axis and translate (x + [-4] , y + [-4])
* You can check the rest of points
# B with H
∵ B = (-3 , 4) after reflection over y-axis is (3 , 4) after translation is
(3 + [-4] , 4 + [-4]) = (-1 , 0) the same with point H = (-1 , 0)
# C with G
∵ C = (-2 , 4) after reflection over y-axis is (2 , 4) after translation is
(2 + [-4] , 4 + [-4]) = (-2 , 0) the same with point G = (-2 , 0)
# D with F
∵ D = (-1 , 2) after reflection over y-axis is (1 , 2) after translation is
(1 + [-4] , 2 + [-4]) = (-3 , -2) the same with point F = (-3 , -2)