Obrazem punktu A(-4;3) w symetrii względem punktu P jest punkt B(2;1). Wynika stąd, że: A.P(1;2) B.P(2;1) C.P(-1;2) D.P(2;-1)
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A(-4;3) B(2;1)
P = (x , y)
x = (-4 + 2)/2 = -2/2 = -1
y = (3 + 1)/2 = 4/2 = 2
odp. P = (-1, 2)