Dany jest odcinek o końcach A=(2, -1) B=(a, 4). Wyznacz a, jeśli odległość AB = 5
5=√((a-2)²+(4-(-1)²))
5=√(a²-4a+4+25)
5=√(a²-4a+29)
25=a²-4a+29
a²-4a+4=0
Δ=(-4)²-4*1*4
Δ=16-16
Δ=0
a=-(-4)/(2*1)
a=4/2
a=2
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5=√((a-2)²+(4-(-1)²))
5=√(a²-4a+4+25)
5=√(a²-4a+29)
25=a²-4a+29
a²-4a+4=0
Δ=(-4)²-4*1*4
Δ=16-16
Δ=0
a=-(-4)/(2*1)
a=4/2
a=2