Question

oblicz pole trójkąta ABCD gdy :

A) A=(8,-4) B=(4,7) C=(1,1)

B) A=(2,2) b=(-2,3) C=(-100,-100)

Answer 1

a)
A =(8;-4), B = (4; 7), C = (1; 1)
  AB = [4 -8; 7-(4)] = [-4 ; 11]
  AC = [1 - 8; 1 -(-4)] = [-7; 5]
I AB I² = (-4)² + 11² = 16 + 121 = 137
 I AB I = √137
I AC I² = (-7)² + 5² = 49 +25 = 74
 I AC I = √74
  AB o AC = -4*(-7) + 11*5 = 28 +55 = 83
 AB o AC = I AB |* I AC I 
 α = [ AB o AC ] : ( I AB i *I AC I )
  α = 83 : ( √137 *√74)
sin²α = 1 - cos² α = 1 - 83²/(137*74)
sin²α = 1 - 6889/10 138 = 3249/10 138
 sinα = 57/√(10 138)
Pole Δ ABC
P = (1/2)*I AB I *I AC I * sin α = (1/2)*√137*√74*[57/√(10 138) =
= (1/2)*57 = 28,5
  b)
A = (2; -2), B = (-2;3), C = (-10; 10)
  AB = [-2-2; 3- (-2)] = [-4; 5]
  AC = [-10 -2; 10 -(-2)] = [-12; 12]
  BC = [ -10 -(-2); 10 -3] = [-8; 7]
 długości: AB, AC, BC :
I AB I² = (-4)² + 5² = 16 +25 = 41
I AC I² = (-12)² + 12² = 144 + 144 = 288
I BC I² = (-8)² + 7² = 64 + 49 = 113
  I AB I = √41
I AC I = 12√2
  AB i AC = (-4)*(-12) + 5*12 = 48 + 60 = 108
 AB i AC = I AB I * I AC I
 108 = √41*12√2* cos α 
 α = 108 : 12√84 = 9/√84
sin² α = 1 - cos²α = 1 - (9/√84)² = 1 - 81/84 = 84/84 - 81/84 =
= 3/84
sin α = √3/ √84
Pole Δ ABC
P = (1/2)*I ABI * I AC I * sin α = (1/2)*√41*12√2*(√3/√84) =
= 6*√246/√84 = 6*√(246/84) = 6*√(41/14) ≈ 10,3