In each diagram, ABC has been transformed to yield A'B'C'. Which transformation could NOT be achieved by rotation alone? A) ABC in quadrant I, A'B'C' in quadrant I under ABC but upside down

B) ABC in quadrant I, A'B'C' in quadrant III but upside down

C) ABC in quadrant I, A'B'C' in quadrant IV but upside down next to ABC both lying on the X axis

D) ABC lying on the y axis between Quadrant I and II, A'B'C' reflection of ABC by the x - axis

E) ABC lying on the y axis between Quadrant I and II, A'B'C' reflection of ABC by the x - axis but the shape is flipped

See the image for a better visual

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

2/3

step-by-step explanation:

the slope of a line is equal to the rise/run. in this case the line rises 2 units for every +3 units it moves