题目:
在平面直角坐标系中,已知A(5,0),B(0,12),且AB=13,在x轴上取一点P,使得△PAB是以AB为腰的等腰三角形,请写出所有符合条件的点P的坐标
(-5,0),(-8,0),(18,0),(-11.9,0)
(-5,0),(-8,0),(18,0),(-11.9,0)
.
答案
(-5,0),(-8,0),(18,0),(-11.9,0)

解:如图,
①若AB=BP,则OA=OP=5,则点P
1(-5,0);
②若AB=AP,则点P
2(-8,0);点P
3(18,0);
③若BP=AP,则cos∠PAB=
=,
∵AC=
AB=
,
∴AP=16.9,
∴OP=AP-OA=11.9,
∴点P
4(-11.9,0).
∴符合条件的点P的坐标分别为:(-5,0),(-8,0),(18,0),(-11.9,0).
故答案为:(-5,0),(-8,0),(18,0),(-11.9,0).