Question

Please please help! I'm stuck on this

(02.01 HC)
Quadrilateral ABCD is located at A(−2, 2), B(−2, 4), C(2, 4), and D(2, 2). The quadrilateral is then transformed using the rule (x − 2, y + 8) to form the image A'B'C'D'. What are the new coordinates of A', B', C', and D'? Describe what characteristics you would find if the corresponding vertices were connected with line segments. (10 points)

Answer 1

A'(-4 , 10)

B'(-4, 12)

C'(0, 12)

D'(0, 10)

Explanation:

Given in the question are 4 coordinates of Quadrilateral ABCD

A(−2, 2)

B(−2, 4)

C(2, 4)

D(2, 2)

The rule is (x − 2, y + 8), that means

the translation is 2 units to the left and 8 units up

Applying the rule of the translation

Step1

A(-2, 2) = A'(-2-2 , 2+8) = A'(-4 , 10)

Step 2

B(−2, 4) = B'(-2-2 , 4+8) = B'(-4, 12)

Step 3

C(2, 4) = C'(2-2 , 4+8) = C'(0, 12)

Step 4

D(2, 2) = D'(2-2 , 2+8) = D'(0, 10)

Describe what characteristics you would find if the corresponding vertices were connected with line segments.

If point A was connected to point A', point B was connected to point B',point C was connected to point C', point D was connected to point D' then all of these line will be parallel to each other, since they were all transformed by the same amount (x-2, y+8).