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Parallelogram JKLM has coordinates J(−8, 16) , K(8, 16) , L(16, −8) , and M(0, −8) . Parallelogram J'K'L'M' has coordinates J′(−1, 2) , K'(1, 2) , L′(2, −1) , and M′(0, −1) . Parallelogram J"K"L"M" has coordinates J"(3, 2), K"(5, 2) , L″(6, −1) , and M″(4, −1) . Which transformations describe why parallelograms JKLM and J"K"L"M" are similar? ​ Parallelogram JKLM ​ was dilated by a scale factor of 1/8 and then translated 4 units to the right. ​ Parallelogram JKLM ​ was dilated by a scale factor of 1/4 and then translated 1 unit right and 2 units down. ​ Parallelogram JKLM ​ was rotated 270° clockwise and then dilated by a scale factor of 1/4 . ​ Parallelogram JKLM ​ was reflected across the y-axis and then dilated by a scale factor of 1/8 .

Verified Answer :

Duhitzamanda128

Sorry i am late, but my answer is for those taking the quiz now,

The correct answer is Parallelogram JKLM ​ was dilated by a scale factor of 1/8 and then translated 4 units to the right. ​

:) Hope this help xd

ashishdwivedilVT

Parallelogram JKLM ​ was rotated 270° clockwise and then dilated by a scale factor of 1/8.

From the distance formula

JK = 16

KL = \(2\sqrt{165}\)

LM = 16

MJ = \(2\sqrt{165}\)

Similarly,

J"K" = 2

K"L" = \(\sqrt{10}\)

L"M"=2

M"J"= \(\sqrt{10}\)

If a parallelogram is reflected about the y-axis then the Y-coordinate of each edge will be unchanged while the X-coordinate will become -X.

so coordinates of parallelogram JKLM after reflection about y-axis will be

J≡(8,16)

K≡(-8,16)

L≡(-16,-8)

M≡(0,-8)

JK = 16

KL = \(2\sqrt{165}\)

LM=16

MJ = \(2\sqrt{165}\)

J"K"/JK = 2/16 = 1/8

it means JKLM was dilated by a factor of 1/8

Hence, Parallelogram JKLM ​ was reflected about the y-axis and then dilated by a scale factor of 1/8.

to get more about dilation refer to:

https://brainly.com/question/10253650

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