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.
+128474 The vertices are ( 7, -7) ( 9,- 7) ( 9, -9) and ( 7, -9)
The point farthest from the origin can be found as
[ old point - dilation center] (scale factor) + dilation center =
[ (9, -9) - (0,0) ] (2) + (0, 0) =
[ 9, - 9 ] (2) + (0, 0) =
(18 , - 18) + (0, 0) =
(18, - 18 )
