Odpowiedź:
A = ( - 10 , 3 ) , B = ( 5 , 3 )
xa = - 10 , xb = 5 , ya = 3 . yb = 3
IABI = √[(xb - xa)² + (yb - ya)²] = √[(5 + 10)² + (3 - 3)²] = √(15² + 0²) =
= √15² = 15
S - środek odcinka IABI = (xs , ys)
xs = (xa + xb)/2 = (- 10 + 5)/2 = - 5/2 = - 2 1/2
ys = (ya + yb)/2 = (3 + 3)/2 = 6/2 = 3
S = (- 2 1/2 ; 3 ) = (- 2,5 ; 3 )
