弦AB=9√10/10
L1:3x+y-6=0,y=6-3x,A(a,6-3a)
L2:3x+y+3=0,y=-3-3x,B(b,-3-3b)
AB^2=(a-b)^2+(9+3b-3a)^2=(9√10/10)^2
(a-b)^2+(9+3b-3a)^2=81/10.(1)
直线L过点C(1,0)
k(AB)=k(AC)=k(BC)
(6-3a)/(a-1)=(-3-3b)/(b-1)
b=3-2a.(2)
(1),(2):
a= b=
A( )
k=
y=k*(x-1)
这题特殊情况的题目,因为两平行直线的距离就=9√10/10
y=(1/3)*(x-1)
L:x-3y-1=0