Answer:
[tex]C'' = (1, -1)[/tex] and [tex]D'' = (0, -5)[/tex]
Step-by-step explanation:
Let's start with C.
[tex](1, 3)[/tex] reflected over the line [tex]y=1[/tex] is [tex](1, 3-2*(3-1))=(1, 3-4)=(1, -1)[/tex]
[tex](1, -1)[/tex] reflected over the line [tex]y=-x[/tex] is the same as flipping the [tex]1[/tex] and [tex]-1[/tex] and taking the negative of both, giving [tex](-(-1), -(1))=(1, -1)[/tex], so [tex]C''=(1, -1)[/tex]
Now, let's do D.
[tex](5, 2)[/tex] reflected over the line y=1 is [tex](5, 2-2*(2-1))=(5, 0)[/tex]
[tex](5, 0)[/tex] reflected over the line y=-x is [tex](-(0), -(5))=(0, -5)[/tex], so [tex]D''=(0, -5)[/tex]
