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ze środka AC:
B = S(AC) = ( (51+x)/2 ; (-18+y)/2 ) = (-14 ; 24 )
(51+x)/2 = -14
51+ x = -28
x = -79
(-18+y)/2 = 24
-18+y = 48
y = 66
odp.: (-79;66)
B=(-14,24)
C= (xC, yC)?
½|AC|=|AB|
Wiemy, że :
[xA+xC]/2 ; [yA+yC]/2=(-14,24)
(51+xC)/2 ; (-18+yC)/2 = (-14,24)
Zatem :
[51+xC]/2=-14
51+xC=-28
xC=-79
[-18+yC]/2=24
-18+yC=48
yC=66