∴设A(a,a),B(b,8b),则有
|
消去m得:(a-b)k=a-8b,
∵当a=b时,a=b=0与题意不符合,
∴a≠b,且k=
| a−8b |
| a−b |
1−
| ||
1−
|
设
| b |
| a |
k=
| 1−8t |
| 1−t |
| 8t−1 |
| t−1 |
| 8(t−1)+7 |
| t−1 |
| 7 |
| t−1 |
∵
| b |
| a |
∴t是整数,且t>0,t≠1;
又∵k为整数,
∴t-1=7或t-1=1,
∴t=8或t=2,
∴k=9或k=15.
| b |
| a |
|
| a−8b |
| a−b |
1−
| ||
1−
|
| b |
| a |
| 1−8t |
| 1−t |
| 8t−1 |
| t−1 |
| 8(t−1)+7 |
| t−1 |
| 7 |
| t−1 |
| b |
| a |