⑴Y=KX+K过(1,4),∴4=K+K,K=2.
∴Y=2X+2,
⑵令X=0,Y=2,令Y=0,X=-1,∴A(-1,0),B(0,2),
OA=1,OB=2,∴AB^2=5,
设直线L交X轴于C,则RTΔABO∽RTΔACB,
∴AB/AO=AC/AB,
∴AB^2=AO*AC=AC=5,∴OC=4,C(4,0),
直线L过B(0,2)与C(4,0)得：Y=-1/2X+2.
⑶设Q(q,0),Y=-1/2X+q,令Y=0,X=2q,
∴AP=|2q+1|,PQ=√5q,AQ=√(1+q^2),
①AP=AQ,4q^2+4q+1=1+q^2,q=-1/4,
AP=3/2,OQ=1/4,SΔAPQ=1/2AP*OQ=3/16,
②PQ=PA,4q^2-4q+1=5q^2,(q+2)^2=5,q=-2±√5,
AP=2√5-3,或AP=2√5-1
SΔAPQ=1/2(2√5-3)(√5-2)=8-9/2√5,
或SΔAPQ=1/2(2√5-1)(√5+2)=4+3/2√5,
③QA=QP,1+q^2=5q^2,q=±1/2,
AP=0(舍去)或2,
SΔAPQ=1/2×2×1/2=1/2.