![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3ac79f3df8dcd10055cf1c90718b4710b9122f34?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:(1)∵直线y=2x+4与x轴交于点A,与y轴交于点B,
∴y=0时,x=-2,x=0时,y=4,故A(-2,0),B(0,4),
由直线AB上有一点Q在第一象限且到y轴的距离为2.
得点Q的横坐标为2,此时y=4+4=8,
所以:Q(2,8);
(2)由A(-2,0)得OA=2
由Q(2,8)可得△APQ中AP边上的高为8,
当点P在x轴的正半轴上时,AP=OA+PO=2+24=26,
S
△APQ=
×26×8=104;
当点P′在x轴的负半轴上时,AP′=P′O-OA=24-2=22,
S
△AP′Q=
×22×8=88.