(1)
y=1
(2)
k(PQ)=-1/2
y-1=(-1/2)*(x+2)
x+2y=0
(3)
y-1=k(x+2)
kx-y+1+2k=0
|-k-2+1+2k|=|3k-0+1+2k|
|k-1|=|5k+1|
k-1=±(5k+1)(这是去绝对值符号呀,这么简单)
k=-1/2,0
y=1
x+2y=0
(1)
y=1
(2)
k(PQ)=-1/2
y-1=(-1/2)*(x+2)
x+2y=0
(3)
y-1=k(x+2)
kx-y+1+2k=0
|-k-2+1+2k|=|3k-0+1+2k|
|k-1|=|5k+1|
k-1=±(5k+1)(这是去绝对值符号呀,这么简单)
k=-1/2,0
y=1
x+2y=0