Dany jest odcinek o końcach A(2,-1) i B(a,4). Jeżeli IABI= 5, to
a) a= -2
b) a=2
c) a=6
d) a= -6
b)
A(2;-1)
B(a;4)
AB=5
AB=√(a-2)²+(4+1)²=√a²-4a+4+25=√a²-4a+29=5
a²-4a+29=5²
a²-4a+29-25=0
a²-4a+4=0
Δ=b²=4ac=16-16=0
a=-b/2a=4/2=2
odp. b
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b)
A(2;-1)
B(a;4)
AB=5
AB=√(a-2)²+(4+1)²=√a²-4a+4+25=√a²-4a+29=5
a²-4a+29=5²
a²-4a+29-25=0
a²-4a+4=0
Δ=b²=4ac=16-16=0
a=-b/2a=4/2=2
odp. b