Odpowiedź:
zad 1
A = ( 3 , - 1 ) , B = ( - 5 , - 2 )
xa = 3 , xb = - 5 , ya = - 1 , yb = - 2
IABI = √[(xb - xa)² + (yb - ya)²] = √[(- 5 - 3)² + ( - 2 + 1)²] = √[(- 8)² + (- 1)²] =
= √(64 + 1) = √65
zad 2
A = ( - 1 , 3 ) , B = ( 3 , 5 )
xa = - 1 , xb = 3 , ya = 3 , yb = 5
S - współrzędne środka odcinka = (xs , ys)
xs = (xa + xb)/2 = ( - 1 + 3)/2 = 2/2 = 1
ys = (ya + yb)/2 = ( 3 + 5)/2 = 8/2 = 4
S = ( 1 , 4 )
