设l 1的解析式为y=k 1x+b 1,
-2 k 1 + b 1 =2
- k 1 + b 1 =0 ,
解得
k 1 =-2
k 2 =-2 ,
∴y=-2x-2.
设l 2的解析式为y=k 2x+b 2,
-2 k 2 + b 2 =2
b 2 =1 ,
解得k 2=-
1
2 ,
∴y=-
1
2 x+1.
故答案为:
y=-2x-2
y=-
1
2 x+1 .
设l 1的解析式为y=k 1x+b 1,
-2 k 1 + b 1 =2
- k 1 + b 1 =0 ,
解得
k 1 =-2
k 2 =-2 ,
∴y=-2x-2.
设l 2的解析式为y=k 2x+b 2,
-2 k 2 + b 2 =2
b 2 =1 ,
解得k 2=-
1
2 ,
∴y=-
1
2 x+1.
故答案为:
y=-2x-2
y=-
1
2 x+1 .