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)?
1 answer
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. :)
2 answers