Question

已知△ABC,(1)∠ACB=900,P为AB边上一动点(不与点A,B重合)过点P引直线截△ABC,

Answer 1

⑴可画三条：如图：

PD⊥AC,PE⊥BC,PF⊥AB,

⑵SΔABC=1/2AC*BC=6,AB=√(AC^2+BC^2)=5,

①PD⊥AC,RTΔAPD∽RTΔABC,

SΔSPD/SΔABC=(AP/AB)^2,

Y=6/25*X^2；

②PE⊥BC,RTΔABC∽RTΔPBE,

Y/6=(5-X)^2/25,

Y=6/25(5-X)^2(或Y=6/25X^2-12/5X+6)

③当F在AC上时(0

PF⊥AB,RTΔAFP∽RTΔABC,

Y/6=(X/4)^2,

Y=3/8X^2.

当F有BC上时(16/5

RTΔBPF∽RTΔBCA,

Y/6=(PB/BC)^2=(5-X)^2/9

Y=2/3(5-X)^2(或Y=2/3X^2-20/3X+50/3.