Đáp án:
Giải thích các bước giải:
(x;y);−−→BC=(−3;3),−−→AH=(x−1;y+1)−−→BH=(x−5;y+3)⇒{AH⊥BC¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯B,H,C⇔{−3(x−1)+3(y+1)=0x−5−3=y+33⇔{3x−3y−6=03x+3y−6=0⇔{x=6y=0⇒H(6;0)H(x;y);BC→=(−3;3),AH→=(x−1;y+1)BH→=(x−5;y+3)⇒{AH⊥BCB,H,C¯⇔{−3(x−1)+3(y+1)=0x−5−3=y+33⇔{3x−3y−6=03x+3y−6=0⇔{x=6y=0⇒H(6;0)