sendo A (0,1) e B (4,5), para quais coordenadas C (x, y) divide AB na razão 1/3
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xC = (xA +1/3*xB)/(1+1/3)
xC = (0 + 1/3*4)/(4/3) = 4/3 / 4/3 = 1
yC = (yA +1/3*yB)/(1+1/3)
yC = (1 + 1/3*5) / 4/3 = 8/3 / 4/3 = 2
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29/11/2015
Sepauto - SSRC
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xC = (0 + 1/3*4)/(4/3) = 4/3 / 4/3 = 1
yC = (yA +1/3*yB)/(1+1/3)
yC = (1 + 1/3*5) / 4/3 = 8/3 / 4/3 = 2
*-*-*-*-*-*-*-*-*-*-*-*
29/11/2015
Sepauto - SSRC
*-*-*-*-*-*-*-*-*-*-*-*
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