数学精品复习资料
规律探索
一、选择?/p>
1.
?/p>
2014?/p>
山东威海?/p>
?/p>
12
?/p>
3
分)
如图?/p>
在平面直角坐标系
xOy
中,
Rt
?/p>
OA
1
C
1
?/p>
Rt
?/p>
OA
2
C
2
?/p>
Rt
?/p>
OA
3
C
3
?/p>
Rt
?/p>
OA
4
C
4
?/p>
的斜边都在坐标轴上,
?/p>
A
1
OC
1
=
?/p>
A
2
OC
2
=
?/p>
A
3
OC
3
=
?/p>
A
4
OC
4
=?30°
?/p>
若点
A
1
的坐标为
?/p>
3
?/p>
0
?/p>
?/p>
OA
1
=
OC
2
?/p>
OA
2
=
OC
3
?/p>
OA
3
=
OC
4
?/p>
,则依此规律,点
A
2014
的纵坐标为(
?/p>
A
?/p>
0
B
?/p>
?/p>
3×
?/p>
?/p>
2013
C
?/p>
?/p>
2
?
2014
D
?/p>
3×
?
?/p>
2013
考点
?/p>
规律型:点的坐标
专题
?/p>
规律型.
分析?/p>
根据?/p>
30
度的直角三角形三边的关系?/p>
OA
2
=
OC
2
=3×
?
OA
3
=
OC
3
=3×
?
?/p>
2
?/p>
OA
4
=
OC
4
=3×
?
?/p>
3
,于是可
得到
OA
2014
=3×
?
?/p>
2013
,由于?/p>
2014=4×
503+2
,则可判?
?/p>
A
2014
?/p>
y
轴的正半轴上,所以点
A
2014
的纵坐标?/p>
3×
?
?
2013
?/p>
解答?/p>
解:∵∠
A
2
OC
2
=30°
?/p>
OA
1
=
OC
2
=3
?/p>
?/p>
OA
2
=
OC
2
=3×
?/p>
?/p>
OA
2
=
OC
3
=3×
?/p>
?/p>
OA
3
=
OC
3
=3×
?
?/p>
2
?/p>
?/p>
OA
3
=
OC
4
=3×
?/p>
?/p>
2
?/p>
?/p>
OA
4
=
OC
4
=3×
?
?/p>
3
?/p>
?/p>
OA
2014
=3×
?
?/p>
2013
?/p>