![](http://hiphotos.baidu.com/zhidao/pic/item/0d338744ebf81a4c949f5d33d42a6059242da69f.jpg)
②當點C在原點與(0,10)之間時,
設C點坐標為(0,y),
則OC=y,AC=
42+y2 |
根據(jù)題意,∠CAO+∠CAB=90°,∠B+∠BAC=90°,
∴∠B=∠CAO,
又∠ACB=∠AOC=90°,
∴△ABC∽△CAO,
∴
CO |
AC |
AC |
AB |
∴AC2=CO?AB,
即42+y2=10y,
∴y2-10y+16=0
解得y1=2,y2=8,
∴點C的坐標為(0,2)(0,8);
故C點的坐標為(0,0)(0,10)(0,2)(0,8).
42+y2 |
CO |
AC |
AC |
AB |