题目
已知直线y=kx-3与两坐标轴围成的三角形面积为6,求k的值.
提问时间:2020-08-03
答案
如图,令y=kx-3=0得x=
,
则直线y=kx-3与x轴交点坐标为(
,0),即A(
,0),
令x=0,得y=-3,则直线y=kx-3与y轴交点坐标为(0,-3)即B(0,-3),
方法1:当k>0时,由S△AOB=
•AO•BO=
•
•3=6,
解得k=
,
当k<0时,由S△AOB=
•AO•BO=
•(-
)•3=6,
解得k=-
,
所以,k=
或k=-
,
方法2:由S△AOB=
•AO•BO=
•|
|•3=6,
解得k=±
.
3 |
k |
则直线y=kx-3与x轴交点坐标为(
3 |
k |
3 |
k |
令x=0,得y=-3,则直线y=kx-3与y轴交点坐标为(0,-3)即B(0,-3),
方法1:当k>0时,由S△AOB=
1 |
2 |
1 |
2 |
3 |
k |
解得k=
3 |
4 |
当k<0时,由S△AOB=
1 |
2 |
1 |
2 |
3 |
k |
解得k=-
3 |
4 |
所以,k=
3 |
4 |
3 |
4 |
方法2:由S△AOB=
1 |
2 |
1 |
2 |
3 |
k |
解得k=±
3 |
4 |
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