设tan
α |
2 |
2x |
1+x2 |
1−x2 |
1+x2 |
2x |
1−x2 |
∵α∈[0°,180°],∴
α |
2 |
∴tan
α |
2 |
∵sinα+cosα=
1 |
5 |
∴
2x |
1+x2 |
1−x2 |
1+x2 |
1 |
5 |
∴6x2-10x-4=0,
解得x=2,或x=-
1 |
3 |
因此tanα=
2x |
1−x2 |
4 |
3 |
∴直线的斜率k=tanα=-
4 |
3 |
故选C
1 |
5 |
4 |
3 |
3 |
4 |
4 |
3 |
3 |
4 |
4 |
3 |
α |
2 |
2x |
1+x2 |
1−x2 |
1+x2 |
2x |
1−x2 |
α |
2 |
α |
2 |
1 |
5 |
2x |
1+x2 |
1−x2 |
1+x2 |
1 |
5 |
1 |
3 |
2x |
1−x2 |
4 |
3 |
4 |
3 |