直线方程 x+(b1)y+c1=0---------(1)
x+(b2)y+c2=0-----------------(2)
(1)x(2) ==>
x^2+b1xy+c1x+b2xy+b1b2y^2+b2cy+c2x+c2b1y+c1c2=0------(3)
kxy+x^2-x+4y-6=0-------------------------------------(4)
比较(3),(4) 系数
xy:b1+b2=k
x:c1+c2=-1
y:b2c1+b1c2=4
c1c2=-6
-6/c2+c2=1
(c2)^2-c2-6=0
(c2-3)(c2+2)=0
c2=3,或 c2=-2
假设c2=3,
c1=-2
-2b2+3b1=4
2b1+2b2=2k
b1=(4+2k)/5
b2=(3k-4)/5
假设c2=-2,
c1=3
b2=(4+2k)/5
b1=(3k-4)/5