Answer:
The vertices of the image will have the coordinates
A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)
B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)
C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)
D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)
Step-by-step explanation:
<em>Note: You missed to mention the vertices of a quadrilateral ABCD. So, I am assuming the following vertices. It would anyways clear your concept.</em>
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Let us suppose
We know that the rule of the dilation with the center of dilation at (2,2) by a scale factor of 2 is:
Therefore, the vertices of the image will have the coordinates
A(-5, 1) → A'(2x-2, 2y-2) → A'(-12, 0)
B(-4, 3) → B'(2x-2, 2y-2) → B'(-10, 4)
C(1, 2) → C'(2x-2, 2y-2) → C'(0, 2)
D(-3, 0) → D'(2x-2, 2y-2) → D'(-8, -2)
