已知A=｛(x,y)∣2x-y=1｝,B=｛(y-1,x+2)∣ax+by=1｝,若A=B則a²+b²=?

已知A=｛(x，y)∣2x－y=1｝，B=｛(y－1，x+2)∣ax+by=1｝，若A=B則a^2+b^2=?
Sol
B=｛(y－1，x+2)∣ax+by=1｝
=｛(m－1，n+2)∣an+bm=1｝
=｛(p，n+2)∣an+b(p+1)=1｝
=｛(p，q)∣a(q－2)+b(p+1)=1｝
=｛(p，q)∣bp+aq+b－2a=1｝
=｛(x，y)∣bx+ay+b－2a=1｝
=｛(x，y)∣bx+ay=1+2a－b｝
So
b/2=a/(－1)=(1+2a－b)/1=w
b=2w
a=－w
1+2a－b=w
1－2w－2w=w
w=1/5
a=－1/5，b=－2/5
a^2+b^2=1/5