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Quadrilateral ABCD has vertices A (-1,8), B (2, 12), C (5,8), and D (-1,-2)

and its image has a translation (x, y) - (x + 12, y-5). What are the new
coordinates of A'B'C'D'?​

User Syko
by
7.4k points

1 Answer

4 votes

Answer:

The new coordinates of A'B'C'D' are
A'(x,y) =(11,3),
B'(x,y) = (14, 7),
C'(x,y) = (17, 3) and
D'(x,y) = (11, -7), respectively.

Explanation:

By using the translation definition, we get the coordinates of the quadrilateral A'B'C'D':


A'(x,y) = A(x,y) + T(x,y) (1)


A'(x,y) = (-1,8) + (12,-5)


A'(x,y) =(11,3)


B'(x,y) = B(x,y) + T(x,y) (2)


B'(x,y) = (2,12) + (12,-5)


B'(x,y) = (14, 7)


C'(x,y) = C(x,y) + T(x,y) (3)


C'(x,y) = (5,8) + (12,-5)


C'(x,y) = (17, 3)


D'(x,y) = D(x,y) +T(x,y) (4)


D'(x,y) = (-1,-2) + (12,-5)


D'(x,y) = (11, -7)

The new coordinates of A'B'C'D' are
A'(x,y) =(11,3),
B'(x,y) = (14, 7),
C'(x,y) = (17, 3) and
D'(x,y) = (11, -7), respectively.

User Steven Cheng
by
7.6k points

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