A triangle has vertices at B(−3, 0), C(2, −1), D(−1, 2). Which transformation would produce an image with vertices B″(1, −2), C″(0, 3), D″(3, 0)? A. (x, y) → (x + 1, y + 1), (x, y) → (y, x)
B. (x, y) → (x + 1, y + 1), (x, y) → (−x, y),
C. (x, y) → (x, −y), (x, y) → (x + 2, y + 2)
D. (x, y) → (−x, y), (x, y) → (x + 2, y + 2)

ithink B.

Step-by-step explanation:

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'.