∵CM⊥l,即kCM?kl=
b+2 |
a?1 |
∴b=-a-1
∴直線l的方程為y-b=x-a,即x-y-2a-1=0
∴|CM|2=(
|1+2?2a?1| | ||
|
∴|MB|2=|CB|2-|CM|2=-2a2+4a+7
∵|MB|=|OM|
∴-2a2+4a+7=a2+b2,得a=-1或
3 |
2 |
當(dāng)a=
3 |
2 |
5 |
2 |
當(dāng)a=-1時(shí),b=0,此時(shí)直線l的方程為x-y+1=0
故這樣的直線l是存在的,方程為x-y-4=0或x-y+1=0.
b+2 |
a?1 |
|1+2?2a?1| | ||
|
3 |
2 |
3 |
2 |
5 |
2 |