Need !

a sequence of transformations maps abc to a’b’c’. the sequence of transformations that maps abc to a’b’c is a reflection across the (blank) followed by a translation (blank)

first blank options
a.) y-axis
b.) x-axis
c.) line y=x
d.) line y=-x

second blank options

a.) 4 units to the right and 10 units up

b.) 8 units to the right and 4 units up

c.) 10 units to the right and 2 units up

d.) 10 units to the right and 4 units up

answer:

the correct answer is " (1) reflect abc across the x-axis and call this new triangle a'b'c'. (2) translate a'b'c' 2 units right and 6 units up so that its image is a''b''c''. "

it is assumed that the points are

a in abc is (1,9), b in abc is (3,12), and c in abc is (4,4).

a'' in a''b''c'' is (3,-3), b'' in a''b''c'' is (5,-6), and c'' in a''b''c'' is (6,2).

both triangles are congruent. since they are congruent, there are no contractions nor dilations occured. after rotating clockwise, we get a"c"b". so we need to reflect it to get a"b"c"

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'.