The image of (triangle) abc is (triangle) a'b'c. what transformations would result in this image? a.(triangle)abc is reflected over the y-axis, then is rotated -90° around the origin. b.(triangle)abc is reflected over the line y = x. c.(triangle)abc is rotated -90° around the origin, then is reflected over the x-axis. d.(triangle)abc is rotated 90° around the origin, then is reflected over y-axis.

The triangle ABC is reflected over the line y = x ⇒ answer B

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

- If point (x , y) reflected across the line y = x

∴ Its image is (y , x)

- If point (x , y) reflected across the line y = -x

∴ Its image is (-y , -x)

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

∴ Its image is (-y , x)

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

∴ Its image is (y , -x)

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

∴ Its image is (-x , -y)

* There is no difference between rotating 180° clockwise or  

anti-clockwise around the origin

* Lets find the vertices of ABC and A'B'C' to solve the problem

- In Δ ABC

# A = (5 , -5) , B = (5 , -4) , C = (2 , -4)

- In Δ A'B'C'

# A' = (-5 , 5) , B = (-4 , 5) , C = (-4 , 2)

∵ The image of (5 , -5) is (-5 , 5)

∵ The image of (5 , -4) is (-4 , 5)

∵ The image of (2 , -4) is (-4 , 2)

∴ The point (x , y) is (y , x)

- From the rule above

∴ The triangle ABC is reflected over the line y = x

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step-by-step explanation: