Răspuns :
ecuatia lui AB: [tex] \frac{X-Xa}{Xb-Xa} = \frac{Y-Ya}{Yb-Ya} [/tex] adica AB= [tex] \frac{x-2}{4-2} = \frac{y-1}{9-1} [/tex] <=> [tex] \frac{x-2}{2} = \frac{y-1}{8} [/tex] = 8x-16=2y-2 <=> 8x-2y-14=0 asa faci si la celelalte laturi.
AH perpendiculara pe BC
panta AH= ( 4-1) /( 3-2)= 3 ; panta AH ·panta BC =-1 ; panta BC= -1/3
dreapta BC : y-9 = -1/3 ·( x-4) ; x +3y -31=0
panta BH = ( 9-4) / ( 4-3) = 5 ; panta BH·panta AC= -1 ; panta AC= -1/5
dreapta AC : y-1 = -1/5·( x-2) ; x+ 5y -7=0
AB : ( x - 2) / ( 4 - 2 ) = ( y - 1 ) /( 9 -1 )
AB :4x -8=y-1
AB: 4x -y -7=0
panta AH= ( 4-1) /( 3-2)= 3 ; panta AH ·panta BC =-1 ; panta BC= -1/3
dreapta BC : y-9 = -1/3 ·( x-4) ; x +3y -31=0
panta BH = ( 9-4) / ( 4-3) = 5 ; panta BH·panta AC= -1 ; panta AC= -1/5
dreapta AC : y-1 = -1/5·( x-2) ; x+ 5y -7=0
AB : ( x - 2) / ( 4 - 2 ) = ( y - 1 ) /( 9 -1 )
AB :4x -8=y-1
AB: 4x -y -7=0