?$解:如圖,易知$? ?$△ABC,△AOC,△BOC為等腰直角三角形$? ?$∴可求得OA=OC=OB=2$? ?$∴A(-2,0),B(2,0)$? ?$C(0,2)$? ?$解:過A作AD⊥x軸于D,BE⊥x軸于E,CF⊥x軸于F,AG⊥BE于G$? ?$∵OA與y軸夾角為30°,AD//y軸,∴∠OAD=30°$? ?$∴可求得OD=1,AD=\sqrt {3}$? ?$不難證得△AOD≌△ABG≌△OCF$? ?$∴BG=1,AG=\sqrt {3},CF=1,OF=\sqrt {3}$? ?$∴A(1,\sqrt {3}),B(1-\sqrt {3},1+\sqrt {3}),C(-\sqrt {3},1)$? ?$解:如圖,分別以AC,BD所在直線建立平面直角坐標(biāo)系$? ?$易求得OA=OB=OC=OD=1$? ?$∴A(0,1),B(-1,0),C(0,-1),D(1,0)$? ?$解:過A作AH⊥BC于H,CK⊥AB于K,FP⊥DE于P$? ?$由題可得,AH=3+1=4,CK=AH=4$? ?$∵△ABC≌△DEF$? ?$∴∠BAC=∠EDF,AC=DF$? ?$在△DPF和△AKC中$? ?${{\begin{cases} {{∠DPF=∠AKC}} \\ {∠PDF=∠KAC} \\ {DF=AC} \end{cases}}}$? ?$∴△DPF≌△AKC(AAS)$? ?$∴FP=CK=4$? ?$即F到y(tǒng)軸距離為4$?
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