Question

Quadrilateral JKLM was dilated according to the rule DO,One-half(x,y)(one-half x, one-half y) to create the image quadrilateral J'K'L'M', which is shown on the graph. On a coordinate plane, quadrilateral J prime K prime L prime M prime has points (0, negative 2), (3, 2), (6, negative 2), and (3, negative 6). What are the coordinates of vertex J of the pre-image?

Answer 1

Given:

Image quadrilateral with points J'(0, -2), K'(3, 2), L'(6, -2) and M'(3, -6)

Rule for the dilation of JKLM to form the image quadrilateral J'K'L'M',

J(x, y) → J'$(\frac{1}{2}x,\frac{1}{2}y)$

To Find:

Coordinates of vertex J of the pre-image.

Solution:

Rule for the dilation from JKLM to J'K'L'M'

J(x, y) → J'$(\frac{1}{2}x,\frac{1}{2}y)$

In other words, rule for the transformation from image to pre-image will be,

J'(x, y) → J(2x, 2y)

J'(0, -2) → J(0, -4)

Therefore, coordinates of point J are (0, -4).

Answer 2

Answer:

(0, -4)

Step-by-step explanation:

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