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?

(0, -4)
(0, -1)
(0, 0)
(0, 4)

Answer 1

Answer:

A. (0,-4)

Step-by-step explanation:

Quadrilateral JKLM was dilated according to the rule

$(x,y)\rightarrow \left(\dfrac{1}{2}x,\dfrac{1}{2}y\right)$

To find the coordinates of the pre-image vertices, you have to use the inverse rule:

$(x,y)\rightarrow (2x,2y)$

On the coordinate plane, quadrilateral J'K'L'M' has vertex J' with coordinates (0,-2), then the coordinates of J are

$(2\cdot 0,2\cdot (-2))=(0,-4)$

Answer 2

Answer:

the guy above deserves brainly

Step-by-step explanation: