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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.

 Dec 7, 2018
 #1
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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 )

 

 

cool cool cool

 Dec 7, 2018

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