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

First find the B coordinate, which is:
[tex]B=(-2,2)[/tex]Now multiply by a scalar of 1/2
[tex]B^{\prime}=\frac{1}{2}(-2,2)=((\frac{1}{2}*-2),(\frac{1}{2}*2))[/tex][tex]B^{\prime}=(-1,1)[/tex]