![](http://hiphotos.baidu.com/zhidao/pic/item/960a304e251f95ca25ce6607ca177f3e6709524c.jpg)
∴OB=3,AB=4
∴0A=
OB2+AB2 |
∴當(dāng)OA為等腰三角形一條腰,則點(diǎn)P的坐標(biāo)是(8,4)(-2,4)(-3,4);
當(dāng)OA為底邊時(shí),
∵A(3,4),
∴直線OA的解析式為y=
4 |
3 |
∴過(guò)線段OA的中點(diǎn)且與直線OA垂直的直線解析式為:y=-
3 |
4 |
25 |
8 |
∴點(diǎn)P的坐標(biāo)是(-
7 |
6 |
故填(8,4)或(-2,4)或(-3,4)或(-
7 |
6 |
OB2+AB2 |
4 |
3 |
3 |
4 |
25 |
8 |
7 |
6 |
7 |
6 |