∴A(4,0),B(0,3),
∴OA=4,OB=3,
∴AB=
32+42 |
(2)依题意BC=t,AC=5-t,AD=t,
若△ACD∽△ABO相似,
![](http://hiphotos.baidu.com/zhidao/pic/item/a8ec8a13632762d09621590ea3ec08fa513dc660.jpg)
∴
AC |
AB |
AD |
AO |
代入得:
5−t |
5 |
t |
4 |
解得:t=
20 |
9 |
∴点C的横坐标也就是AO-AD=AO-t=4-
20 |
9 |
16 |
9 |
再把x=
16 |
9 |
5 |
3 |
∴此时C(
16 |
9 |
5 |
3 |
若△ACD∽△AOB相似,
AD |
AB |
AC |
AO |
t |
5 |
5−t |
4 |
∴t=
25 |
9 |
AC=5-t=
20 |
9 |
再过C点做CE⊥OA于E,
然后△ACE∽ABO,
AE |
AO |
AC |
AB |
即
AE |
4 |
| ||
5 |
解得AE=
16 |
9 |
∴OE=AO-AE=4-
16 |
9 |
20 |
9 |
而且又∵
CE |
OB |
AE |
AO |
CE |
3 |
| ||
4 |
解得CE=
4 |
3 |
20 |
9 |
4 |
3 |
∴C(
16 |
9 |
5 |
3 |
20 |
9 |
4 |
3 |