Punkt S(-8,b) jest środkiem odcinka AB, gdzie A(3a,2), B(a,3b). Oblicz a i b.
S= (x,y) A=(x,y) B=(x,y)
x :
(3a+a)/2=-8 /*2
4a = -16 /:4
a = -4
y:
(2 + 3b)/2 = b /*2
2 + 3b = 2b
3b - 2b = -2
b = -2
odp:
a=-4 , b=-2
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S= (x,y) A=(x,y) B=(x,y)
x :
(3a+a)/2=-8 /*2
4a = -16 /:4
a = -4
y:
(2 + 3b)/2 = b /*2
2 + 3b = 2b
3b - 2b = -2
b = -2
odp:
a=-4 , b=-2