Answer:
J'(2, -2)
K'(-4, 4)
L'(-1, 2)
Step-by-step explanation:
To reflect the points J(2, 4), K(-4, -2), and L(-1, 0) across the line [tex]y = 1[/tex], we will use the reflection formula:
[tex](x, y) \to (x, 2k - y)[/tex]
where
[tex]k = 1[/tex] (the [tex]y[/tex]-coordinate of the line of reflection).
Reflecting point J(2, 4):
[tex] J'(2, 4) \to (2, 2(1) - 4) = (2, 2 - 4) = (2, -2) [/tex]
So, the reflection of J(2, 4) across [tex]y = 1[/tex] is J'(2, -2).
Reflecting point K(-4, -2):
[tex] K'(-4, -2) \to (-4, 2(1) - (-2)) = (-4, 2 + 2) = (-4, 4) [/tex]
So, the reflection of K(-4, -2) across [tex]y = 1[/tex] is K'(-4, 4).
Reflecting point L(-1, 0):
[tex] L'(-1, 0) \to (-1, 2(1) - 0) = (-1, 2) [/tex]
So, the reflection of L(-1, 0) across [tex]y = 1[/tex] is L'(-1, 2).
Therefore, the reflected points across the line [tex]y = 1[/tex] are:
J'(2, -2)
K'(-4, 4)
L'(-1, 2)
Note: J' is A, K' is B and L' is C in graph.
