goiduongtronCcotamI(1;−1)vabankinhR=5.khoangcachtuIdenAla√52.goiHlachanduongcaoketuIdenBC.⇔AC=CH=BH.⇔AH=2ACcoAC.AB=AC.3AC=3AC2.coAI2−R2=AH2+IH2−R2=4AC2−(R2−IH2)=4AC2−AC2=3AC2⇔AC.AB=AI2−R2.
⇔3AC2=52−25=27⇔AC2=9⇔AC=3.⇔IH=√52−32=4.
⇔(I,AB)=4vaABdiquaA(7;3)⇔ptduongthangcantimla:y=3hoac12x−5y−69=0