如图,已知一次函数y=-x+7与正比例函数y= x的图象交于点A,且与x轴交于点B.

（1）解得A(3,4),B(7,0),C(0,4),直线l的解析式为x=7当0≤t≤4时,P(0,t),R(7-t,0),Q(7-t,t)直线PR解析式为y=[t/(t-7)]x+t作AH⊥x轴,垂足H,AH交PR于M,作PN⊥AH,垂足N则M(3,(t^2-4t)/(t-7)),N(3,t),H(3,0)∴AM=4-(t^2-4t)/(t-7)=(-t^2+8t-28)/(t-7)PN=3,RH=4-t故S△APR=S△APM+S△ARM=(1/2)AM*PN+(1/2)AM*RH=(1/2)[(-t^2+8t-28)/(t-7)]*[3+4-t]=0。5t^2-4t+14令其等于8解得t=2或t=6,后者舍去当4≤t≤7时,P(t-4,4),R(7-t,0),Q(7-t,(28-4t)/3),作RH'⊥AC,垂足H'则H'(7-t,4)∴AP=7-t,RH=4∴S△APR=(1/2)AP*RH=14-2t,令其等于8解得t=3舍去综上所述,t=2时S△APR面积为8（2）当0≤t≤4时,P(0,t),R(7-t,0),Q(7-t,t),可见PQ//x轴此时欲使△APQ为等腰三角形,显然只有使AP=AQ则Q的横坐标应为(3-0)x2=6,即7-t=6得t=1当4≤t≤7时,P(t-4,4),R(7-t,0),Q(7-t,(28-4t)/3),AP=7-t,AQ=(5t-20)/3,PQ=√[(2t-11)^2 +(4-(28-4t)/3)^2]=(√(52t^2 -524t+1345))/3若AP=AQ则t=41/8若AP=PQ则解得t=226/43或t=4（t=4时Q与A重合,故此情况舍去）若AQ=PQ则解得t=5或t=7（t=7时A与P重合,故此情况舍去）综上所述,满足要求的t为：t=41/8或t=226/43或t=5