Odpowiedź:
a)
A = (-5, -4 ) , B = ( 2 , - 4 ) , C = ( 2 , 3 ) , D = ( - 5 , 3 )
xa = - 5 , xb = 2 , xc = 2 , xd = - 5
ya = - 4 , yb = - 4 , yc = 3 , yd = 3
Ponieważ xa = xd i xb = xc oraz ya = yb i yc = yd , więc podany czworokąt jest kwadratem
a - jeden bok kwadratu = IxaI + xb= I - 5I + 2 = 5 + 2 = 7 [j]
b - drugi bok kwadratu = IyaI + yd = I - 4I + 3 = 4 + 3 = 7 [j]
P - pole = a * b = 7 * 7 = 49 [j²]
[j] - znaczy właściwa jednostka
b)
A = ( -2,4) , B = ( 4, 4) , C= (0,2 )
xa = - 2, xb = 4 , xc = 0 , ya = 4 , yb = 4 , yc = 2
P = 1/2I(xb - xa)(yc - ya) - (yb - ya)(xc - xa)I =
= 1/2 I(4 + 2)(2 - 4) - (4 - 4)(0 + 2)I = 1/2I6 * (- 2) - 0 * 2)I =
= 1/2I (- 12)- 0I = 1/2I- 12) = 1/2 * 12 = 6 [j²]
c)
A = ( - 5 , 3 ) , B = ( 1 , 3 ) , C = ( 0 , 8 )
xa = - 5 , xb = 1 , xc = 0 , ya = 3 , yb = 3 , yc = 8
P = 1/2I(xb - xa)(yc - ya) - (yb - ya)(xc - xa)I =
= 1/2 I(1+ 5)(8 - 3) - ( 3 - 3)(0 + 5)I = 1/2I(6 * 5 - 0 * 5I =
= 1/2I30 - 0I = 1/2I30I = 1/2 * 30 = 15 [j²]
