Asked by morgan
△DEF has vertices D (2,2), E (-2,-1), and F (-3,5). Complete the following charts indicating the location of △D′E′F′ and △D′′E′′F′′
△DEF
is reflected in the y-axis. Then it is translated along the vector ⟨3,−5⟩.
D (2,2) D`E`F` (?,?), D"E"F" (?,?)
E (-2,1) D`E`F` (?,?) D"E"F" (?,?)
F (-3,5) D`E`F` (?,?) D"E"F" (?,?)
△DEF
is reflected in the y-axis. Then it is translated along the vector ⟨3,−5⟩.
D (2,2) D`E`F` (?,?), D"E"F" (?,?)
E (-2,1) D`E`F` (?,?) D"E"F" (?,?)
F (-3,5) D`E`F` (?,?) D"E"F" (?,?)
Answers
Answered by
MathMate
Given:
D (2,2), E (-2,-1), and F (-3,5).
The reflection about the y-axis follows the following rule:
sy: (x,y)->(-x,y)
For example:
sy: D(2,2)->D'(-2,2)
The translation along a vector (p,q) follows the following rule:
T(p,q): (x,y)->(x+p, y+q)
Example:
T(3,-5): D'(-2,2)-> D"(-2+3,2-5)=D"(1,-3)
You can proceed in a similar way for E and F.
Post your results for checking if you wish.
D (2,2), E (-2,-1), and F (-3,5).
The reflection about the y-axis follows the following rule:
sy: (x,y)->(-x,y)
For example:
sy: D(2,2)->D'(-2,2)
The translation along a vector (p,q) follows the following rule:
T(p,q): (x,y)->(x+p, y+q)
Example:
T(3,-5): D'(-2,2)-> D"(-2+3,2-5)=D"(1,-3)
You can proceed in a similar way for E and F.
Post your results for checking if you wish.
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