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