$ 解:(1)由題,3a-11-2=0,5=2b-1,解得a=\frac {13}{3},b=3$ $(2)由題,3a-11-3=-2,5+2b-1=0$ $解得a=4,b=-2$ $∴±\sqrt {a+b}=±\sqrt {4-2}=±\sqrt {2}$
$解:(2)A_{1}(-2,2),B_{1}(-3,0),C_{1}(0,0)$ $(3)周長為3+\sqrt {1^{2}+2^{2}}+\sqrt {2^{2}+2^{2}}=3+\sqrt {5}+2\sqrt {2}$ $面積為\frac {1}{2}×3×2=3$ $(2)解:將△ABC先向左平移5個(gè)單位,再向上平移4個(gè)單位即可得到△A'B'C'$ $(3)結(jié)合(2)可知,m-5=2m-8,4-n+4=n-4$ $解得m=3,n=6$
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