Question

"Quadrilateral ABCD underwent a sequence of transformations to give quadrilateral A′B′C′D′. Which transformations could have taken place?
a reflection across the x-axis followed by a reflection across the y-axis

a translation 10 units down followed by a translation 8 units to the right

a rotation 90° counterclockwise about the origin followed by a reflection across the y-axis

a reflection across the line y = x followed by a reflection across the line y = -x"

Answer 1

The quadrilateral ABCD have vertices at points A(-6,4), B(-6,6), C(-2,6) and D(-4,4).

Translating  10 units down you get points A''(-6,-6), B''(-6,-4), C''(-2,-4) and D''(-4,-6).

Translaitng 8 units to the right you get points A'(2,-6), B'(2,-4), C'(6,-4) and D'(4,-6) that are exactly vertices of quadrilateral A'B'C'D'.

Answer: correct choice is B.

Answer 2

Answer:

The transformation that had taken place to map Quadrilateral ABCD to Quadrilateral A'B'C'D' is:

     A translation 10 units down followed by a translation 8 units    to the right

Step-by-step explanation:

We are given a vertices of Quadrilateral ABCD as:

A(-6,4) , B(-6,6) , C(-2,6) , D(-4,4)

Now when translation is  done 10 units down  then the rule that follows this transformation is:

                 (x,y) → (x,y-10)

and coordinate of the transformed image will be:

A(-6,4) → A°(-6,-6)

B(-6,6) → B°(-6,-4)

C(-2,6) → C°(-2,-4)

D(-4,4) → D°(-4,-6)

Now when this image is translated 8 units to the right we obtain image A'B'C'D' and the coordinates of the transformed image follows the rule:

                       (x,y) → (x+8,y)

Hence,

A°(-6,-6) → A'(2,-6)

B°(-6,-4) → B'(2,-4)

C°(-2,-4) → C'(6,-4)

D°(-4,-6) → D'(4,-6)

Hence,

A(-6,4) → A'(2,-6)

B(-6,6) → B'(2,-4)

C(-2,6) → C'(6,-4)

D(-4,4) → D'(4,-6)