(1)如图1,过点B作BC⊥OA,垂足为C
∵△OAB为等边三角形,A的坐标(2,0)
∴BO=OA=2,OC=1,∠BOC=60°,
∴BC=
.
∴B的坐标
(1,).
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/42166d224f4a20a484147c3d93529822730ed0c6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
(2)∵△OAB与△APQ为等边三角形
∴∠BAO=∠PAQ=60°
∴∠BAQ=∠OAP,
在△APO和△AQB中,
∴△APO≌△AQB(SAS),
∴∠ABQ=∠AOP=90°,
∴当点P在x轴上运动(P不与O重合)时,∠ABQ为定值90°;
(3)解:由(2)可知,点Q总在过点B且与AB垂直的直线上,可见AO与BQ不平行.
①如图2,当点P在y轴负半轴上时,点Q在点B的左方,
此时,若AB∥OQ,四边形AOQB即是梯形,
当AB∥OQ时,∠BQO=90°,∠BOQ=∠ABO=60°.
又∵OB=OA=2,可求得BQ=
,由(2)可知,△APO≌△AQB,
∴OP=BQ=
,
∴此时P的坐标为(0,-
).
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d8f9d72a6059252d0d1d5dfe379b033b5ab5b9c6?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
②如图3,当点P在y轴正半轴上时,点Q在B的上方,
此时,若AQ∥OB,四边形AOBQ即是梯形,
当AQ∥OB时,∠ABQ=90°,∠QAB=∠ABO=60°.
又∵AB=2,可求得BQ=
2,
由(2)可知,△APO≌△AQB,
∴OP=BQ=
2,
∴此时P的坐标为((0,2
).
综上,P的坐标为P的坐标为(0,-
)或者为(0,2
).
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/7aec54e736d12f2eb32dee9a4cc2d5628535681b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)