In the coordinate plane, quadrilateral ABCD has vertices A(-2,3) B(4,5) C(10,1) and D(8,-9). Let E, F,G, and H be the midpoints of AB, BC, CD and DA, respectively. Prove that quadrilateral EFGH is a parallelogram

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.

1

x = 1

step-by-step explanation:

0 ÷ x = 0

0 ÷ 1 = 0

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)