The preimage of square ABCD has its center at (8,-8) and has an area of 4 square units. The top side of the square is horizontal. The square is then dilated with the dilation center at (0,0) and a scale factor of 2. What are the coordinates of the vertex of the image of square ABCD that is farthest from the origin? Give your answer as an ordered pair.

I don't fully understand this problem. How does it get dilated at the origin when the square ABC had center at (8,-8)?

I apologize for any confusion. Let me clarify the process for you.

The preimage of square ABCD has its center at (8, -8) and its top side is horizontal. This means that the original square ABCD is positioned with its center at (8, -8) and its vertices are located relative to this center.

When the square is dilated with a dilation center at (0, 0) and a scale factor of 2, it means that every point of the original square is multiplied by the scale factor of 2, with the dilation center as the reference point.

To find the vertex of the image that is farthest from the origin, we need to determine which vertex of the original square is farthest from the origin (0, 0). Since the top side of the square is horizontal and the center of the original square is (8, -8), we can determine that the top right vertex of the original square is farthest from the origin.

The coordinates of the top right vertex of the original square are (8 + side length/2, -8 - side length/2).

Now, let's calculate the side length of the original square. We know that the area of the original square is 4 square units. Since a square has equal side lengths, the side length of the original square is √4 = 2 units.

Substituting the side length into the formula for the top right vertex, we get:
(8 + 2/2, -8 - 2/2) = (9, -9)

Therefore, the coordinates of the vertex of the image of square ABCD that is farthest from the origin after dilatation are (9, -9).

I understand your confusion. Let me explain how the dilation works in this problem.

The initial square, ABCD, has its center at (8, -8). This means that the distance from the center of the square to any of its vertex points is half the length of one side of the square. Since the square has an area of 4 square units, the length of one side is 2 units.

Now, when a figure is dilated, each point of the figure is moved away or closer to a center point, called the dilation center. In this case, the dilation center is at (0, 0).

The scale factor of 2 means that the length of each side of the original square will double in the image square. So, the new side length is 2 * 2 = 4 units.

To find the coordinates of the vertex of the image square farthest from the origin, we need to consider the original square's coordinates and apply the dilation.

The top side of the original square was horizontal, which means that AB and CD are parallel to the x-axis. Therefore, the coordinates of A and B have a y-coordinate of -8, and the coordinates of C and D have a y-coordinate of -8 + 2 = -6 (since the length of one side doubled).

Applying the dilation, we calculate the new coordinates of the vertex farthest from the origin.

For A: (8, -8) -> (8 * 2, -8 * 2) = (16, -16)
For B: (8 + 2, -8) -> (10 * 2, -8 * 2) = (20, -16)
For C: (8 + 2, -6) -> (10 * 2, -6 * 2) = (20, -12)
For D: (8, -6) -> (8 * 2, -6 * 2) = (16, -12)

Out of these coordinates, the vertex with the farthest distance from the origin is (20, -16). Therefore, the answer is (20, -16).

I hope this explanation helps you understand how the dilation is applied and how to find the coordinates of the vertex farthest from the origin in the image square.

Enter keyword "Dilation Center" into Google and look at the "Math Open Reference" website for information. Try moving the center of dilation outside the rectangle they have there. You will then see the larger rectangle move as well. The dilation point determines where your larger, dilated figure will be. Now click "Show distances". The distance from the origin to any point on the small rectangle, doubled (because our scale factor is 2), would be the distance from the origin to the corresponding point on the larger, dilated rectangle. The vertex of our SMALLER square that is farthest from the origin would be (9,-9). So the distance from that point to the origin is 9 * square root of 2. That distance doubled would be 18 * square root of 2. The end point of this line segment has coordinates (18,-18). Notice because the length from the origin was doubled, the coordinates were doubled as well. :)