Question

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'?​

Answer 1

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.

Step-by-step 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.