B分别与OA、OB交于A、B.(Ⅰ)当AB的中点为P时,求直线AB的方程;(Ⅱ)当AB的中点在直线xy21?上时,求直线AB的方程.AA13.05???yx,023??yx14.0162???yx15.30°16.]5,4[?17.105°;165°18.1319.07???yx和0223???yx.20.(Ⅰ)32h?,2 21 3 63 3( )3 4 8V h a ab b? ?????.(Ⅱ)3h?,39'2h?,1 27 39 27 39(3 3 ) '2 2 2 4S a b h? ????.21.由???????0120yxy得?????01yx,即A的坐标为)0,1(?,∴1102???ABk,又∵x轴为∠BAC的平分线,∴1????ABACkk,又∵直线012???yx为BC边上的高,∴2??BCk.设C的坐标为),(ba,则11???ab,212????ab,解得5?a,6?b,即C的坐标为)6,5(.22.(Ⅰ)MO//AC1;(Ⅱ)MO∥AC1,AC1⊥平面D1B1C,MO⊥平面D1B1C,平面D1B1C⊥平面B1MC.23.解:(Ⅰ)由题意得,OA的方程为xy?,OB的方程为xy33??,设),(aaA,),3(bbB?。∵AB的中点为)0,1(P,∴???????023baba得13??a,∴132313??????ABk即AB方程为013)13(?????yx(Ⅱ)AB中点坐标为)2,23(baba??在直线xy21?上,则23212baba????,即ba)32(???①∵PBPAkk?,∴131????bbaa②由①、②得3?a,则233??ABk,所以所求AB的方程为0332)33(?????yx
全文预览