Coordinate plane with quadrilaterals EFGH and E prime F prime G prime H prime at E 0 comma 1, F 1 comma 1, G 2 comma 0, H 0 comma 0, E prime

Question

Coordinate plane with quadrilaterals EFGH and E prime F prime G prime H prime at E 0 comma 1, F 1 comma 1, G 2 comma 0, H 0 comma 0, E prime negative 1 comma 2, F prime 1 comma 2, G prime 3 comma 0, and H prime negative 1 comma 0. F and H are connected by a segment, and F prime and H prime are also connected by a segment. Quadrilateral EFGH was dilated by a scale factor of 2 from the center (1, 0) to create E’F’G’H’. Which characteristic of dilations compares segment F’H’ to segment FH

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Khải Quang 1 week 2021-07-22T22:21:22+00:00 1 Answers 5 views 0

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    2021-07-22T22:23:19+00:00

    Answer:

    |F'H'| = 2 * |FH|

    Step-by-step explanation:

    Given

    E = (0,1)             E' = (-1,2)

    F = (1,1)             F' = (1,2)

    G = (2,0)             G' =(3,0)

    H = (0,0)            H' = (-1,0)

    (x,y) = (1,0) — center

    k = 2 — scale factor

    See comment for proper format of question

    Required

    Compare FH to F’H’

    From the question, we understand that the scale of dilation from EFGH to E’F’G’H is 2;

    Irrespective of the center of dilation, the distance between corresponding segment will maintain the scale of dilation.

    i.e.

    |F'H'| = k * |FH|

    |F'H'| = 2 * |FH|

    To prove this;

    Calculate distance of segments FH and F’H’ using:

    d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

    Given that:

    F = (1,1)             F' = (1,2)

    H = (0,0)            H' = (-1,0)

    We have:

    FH = \sqrt{(1- 0)^2 + (1- 0)^2}

    FH = \sqrt{(1)^2 + (1)^2}

    FH = \sqrt{1 + 1}

    FH = \sqrt{2}

    Similarly;

    F'H' = \sqrt{(1 --1)^2 + (2 -0)^2}

    F'H' = \sqrt{(2)^2 + (2)^2}

    Distribute

    F'H' = \sqrt{(2)^2(1 +1)}

    F'H' = \sqrt{(2)^2*2}

    Split

    F'H' = \sqrt{(2)^2} *\sqrt{2}

    F'H' = 2 *\sqrt{2}

    F'H' = 2\sqrt{2}

    Recall that:

    |F'H'| = 2 * |FH|

    So, we have:

    2\sqrt 2 = 2 * \sqrt 2

    2\sqrt 2 = 2\sqrt 2 — true

    Hence, the dilation relationship between FH and F’H’ is::

    |F'H'| = 2 * |FH|

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