1.
(1) L1⊥L2 ==> 两斜率为负倒数,k1*k2=-1,==> (a/b)*(-1/(a-1))=-1 .
L1过点（-3,-1）==> -3a+b+4=0 .
联立得到,{a = 2/3,b = -2},或{a = 2,b = 2}.
（2）L1‖L2 ==> 斜率相等,k1=k2 ==> (a/b)=(-1/(a-1)) .
点到直线距离d=|f(x0,y0)|/sqrt(A^2+B^2),
==> 4/sqrt(a^2+b^2)=|b|/sqrt[(a-1)^2+1].
联立得到 a = 2,b = -2.
2.
(1) 点斜式,BC; y-1=(3-1)/(-2-2)*(x-2)=-(x-2)/2 ==>x+2y-4=0.
(2) 点D(0,2),AD; y-2=(0-2)/(-3-0)*(x-0)=2x/3.
(3) DE; y-2=2x.
3.
(1) A'(x1,y1),依题意,直线L是AA'中垂线,
故AA'中点((x1-1)/2,(y1-2)/2)在直线L上,
有 x1-1-3(y1-2)/2+1=0,
并且 斜率互为负倒数,(y1+2)/(x1+1)=-3/2.
联立得到 x1=3/2,y1=-23/4.
(2)
在线m上取点B(2,0),则B关于L的对称点B'(-3/2,21/4).(类似（1）求得)
而直线L和m相交点C(4,3),
故直线B'C,y-3=(3-21/4)/(4+3/2)*(x-4) ==> 9x+22y=102.
即为所求.