Triangle xyz with vertices x(0, 0), y(0, –2), and z(–2, –2) is rotated to create the image triangle x'(0, 0),
y'(2, 0), and z'(2, –2).

which rules could describe the rotation? check all that apply.

r0, 90°
r0, 180°
r0, 270°
(x, y) → (–y, x)

(x, y) → (y, –x)

The answer is R0, 90° and (x , y) → (-y , x) ⇒1st and 4th

Step-by-step explanation:

* Lets revise the rotation rules

- If point (x , y) rotated about the origin by angle 90° anti-clock wise

 (+90° or -270°)

∴ Its image is (-y , x)

- If point (x , y) rotated about the origin by angle 90° clock wise

 (-90° or +270°)

∴ Its image is (y , -x)

- If point (x , y) rotated about the origin by angle 180°

 (+180° or -180°)

∴ Its image is (-x , -y)

* There is no difference between rotating 180° clockwise or  

  anti-clockwise around the origin

* Lets solve the problem

∵ Δ XYZ has vertices⇒ X (0 , 0) , Y (0 , -2) , Z (-2 , -2)

∵ Δ X'Y'Z' has vertices X' (0 , 0) , Y' (2 , 0) , Z' (2 , -2)

* From them

# Y = (0 , -2) and Y' = (2 , 0), that means the image is (-y , x)

# Z = (-2 , -2) and Z' = (2 , -2), that means the image is (-y , x)

∴ The rotation is around the origin with angle 90° anti-clockwise

V.I.N: Anti-clock wise means positive angle , clockwise means

negative angle (90° means anti-clockwise , -90° means clockwise)

∴ The answer is: R0, 90° and (x , y) → (-y , x)

the answer is 6

hope this !