Answer:
a) C'(-1,2), D'(1,7), E(-2,6) and (-4,1)
b) C''(-4,-5), D''(-2,0), E''(-5,-1), F'(-7-6)
Step-by-step explanation:
a)
We want to find the coordinates of Parallelogram CDEF with vertices C(2, -1), D(7,1), E(6,-2), and F(1,-4) after a reflection across y = x.
We can find these coordinates by following the following rule:
Reflection over y = x rule : ( x , y ) ---> ( y , x )
Explanation of rule : the x and y values simply swap places .
Applying rule to coordinates:
C(2,-1) --- swap 2 and - 1 ---> C'(-1,2)
D(7,1) --- swap 7 and 1 ---> D'(1,7)
E(6,-2) --- swap 6 and -2 ---> E'(-2,6)
F(1,-4) --- swap 1 and -4 ---> F'(-4,1)
The coordinates of Parallelogram CDEF after a reflection across y = x are C'(-1,2), D'(1,7), E(-2,6) and (-4,1)
For more validation check the attached image.
b)
We now take the coordinates of the parallelogram after the reflection and translate it with a rule of (x, y) – (x – 3, y-7)
Explanation of rule: Subtract 3 from the x value and subtract 7 from the y value
Evaluation of translation:
C'(-1,2) ---> (-1 - 3 = -4 , 2 - 7 = -5 ) ---> C''(-4,-5)
D'(1,7) ---> (1 - 3 = -2 . 7 - 7 = 0 ) ---> D''(-2,0)
E'(-2,6) ---> (-2 - 3 = -5 , 6 - 7 = -1 ) ---> E''(-5,-1)
F'(-4,1) ---> (-4 - 3 = -7 , 1 - 7 = -6 ) ---> F''(-7,-6)
The coordinates of the parallelogram after a translation with a rule of (x, y) – (x – 3, y-7) are C''(-4,-5), D''(-2,0), E''(-5,-1), F'(-7-6)
For more validation check the other attached image
