k |
x |
∴1=
k |
2 |
解得:k=2,
∵点P(2,1)在一次函数y=kx+b的图象上,
∴1=2×2+b,
解得:b=-3,
∴k=2,b=-3;
(2)图象如右图:
(3)∵E(
3 |
2 |
∴S△EOF=
1 |
2 |
1 |
2 |
3 |
2 |
9 |
4 |
S△EOP=
1 |
2 |
3 |
2 |
3 |
4 |
∴S△EOF=3S△EOP;
(4)能.理由如下:
若S△QOE=S△EOF,
则Q的纵坐标为±3,
令y=±3,代入y=
k |
x |
2 |
3 |
∴Q(
2 |
3 |
2 |
3 |
k |
x |
k |
x |
k |
x |
k |
2 |
3 |
2 |
1 |
2 |
1 |
2 |
3 |
2 |
9 |
4 |
1 |
2 |
3 |
2 |
3 |
4 |
k |
x |
2 |
3 |
2 |
3 |
2 |
3 |