If ABCD is dilatedby a factor of 1/2, thecoordinate of B' would be:B'= ([?],[ ])

asked Apr 23, 2023 87.2k views

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First find the B coordinate, which is:

Now multiply by a scalar of 1/2

$B^(\prime)=(1)/(2)(-2,2)=(((1)/(2)*-2),((1)/(2)*2))$
$B^(\prime)=(-1,1)$

Cameron Walsh answered Apr 30, 2023

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