如圖,已知一次函數(shù)y=?
x+3的圖象與x軸,y軸分別相交于A,B兩點(diǎn),點(diǎn)C在AB上以
![](http://hiphotos.baidu.com/zhidao/pic/item/aa64034f78f0f7368233beb00955b319eac413b5.jpg)
每秒1個(gè)單位的速度從點(diǎn)B向點(diǎn)A運(yùn)動(dòng),同時(shí)點(diǎn)D在線段AO上以同樣的速度從點(diǎn)A向點(diǎn)O運(yùn)動(dòng),運(yùn)動(dòng)時(shí)間用t(單位:秒)表示.
(1)求AB的長(zhǎng);
(2)當(dāng)t為何值時(shí),△ACD與△ABO相似?并直接寫(xiě)出此時(shí)點(diǎn)C的坐標(biāo).
(1)當(dāng)x=0時(shí),y=3;當(dāng)y=0時(shí),x=4,
∴A(4,0),B(0,3),
∴OA=4,OB=3,
∴AB=
=5;
(2)依題意BC=t,AC=5-t,AD=t,
若△ACD∽△ABO相似,
![](http://hiphotos.baidu.com/zhidao/pic/item/a8ec8a13632762d09621590ea3ec08fa513dc660.jpg)
∴
=
,
代入得:
=
,
解得:t=
,
∴點(diǎn)C的橫坐標(biāo)也就是AO-AD=AO-t=4-
=
,
再把x=
帶入一次函數(shù)解析式,得y=
.
∴此時(shí)C(
,
)
若△ACD∽△AOB相似,
=
,
=
,
∴t=
,
AC=5-t=
,
再過(guò)C點(diǎn)做CE⊥OA于E,
然后△ACE∽ABO,
=
,
即
=
,
解得AE=
,
∴OE=AO-AE=4-
=
,
而且又∵
=
,即
=
.
解得CE=
.所以C(
,
)
∴C(
,
)或(
,
)