Odpowiedź:
A = ( 1 , - 1 ) , B = ( - 3 , - 4) , C = ( 5 , 7 )
xa = 1 , xb = - 3 , xc = 5 , ya = - 1 , yb = - 4 , yc = 7
P - pole trójkąta = 1/2I(xb - xa)(yc - ya) - (yb - ya)(xc - xa)I =
= 1/2I(- 3 - 1)(7 + 1) - ( - 4 + 1)(5 - 1)I = 1/2I(- 4) * 8 - (- 3) * 4I =
= 1/2I - 32 - (- 12I = 1/2I - 32 + 12I = 1/2I- 20I = 1/2 * 20 = 10
IABI = √[(xb - xa)² + (yb - ya)²] = √[(- 3 - 1)² + ( - 4+1)²] = √[(- 4)²+ (- 3)²] =
= √(16 + 9) = √25 = 5
IACI = √[(xc - xa)² + (yc - ya)²] = √[(5 - 1)² + (7 + 1)²] = √(4² + 8²) =
= √(16 + 64) = √80 = √(16 * 5) = 4√5
IBCI = √[(xc - xb)² + (yc - yb)²] = √[(5 + 3)² + (7 + 4)²] = √(8² + 11²) =
= √(64 + 121) = √185
h₁ =P : IABI = 10 : 5 = 2
h₂ = P : IACI = 10 : 4√5 = 10√5/(4 * 5) = 10√5/20 = √5/2
h₃ = P : IBCI = 10 : √185 = 10√185/185 = 2√185/37
