Obter um ponto P no eixo das abcissas equidistantes dos pontos A(2,-3,1) e B(-2,1,-1)
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P(x,0,0), A(2,-3,1), C(-2,1,-1)
dPA = dPB
dPA = √[(xP-xA)² + (yP-yA)² + (zP-zA)²] = √[10+(-2+x)^2]
dPB = √[2+(2+x)^2]
{√[10+(-2+x)^2]}² = {√[2+(2+x)^2]}²
[10+(-2+x)^2] = [2+(2+x)^2]
10 + 4 - 4x + x² = 2 + 4 + 4x + x²
-4x - 4x = -14 + 6
-8x = -8
x = -8/-8
x = 1
P = (1,0,0)
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08/04/2016
Sepauto - SSRC
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dPA = dPB
dPA = √[(xP-xA)² + (yP-yA)² + (zP-zA)²] = √[10+(-2+x)^2]
dPB = √[2+(2+x)^2]
{√[10+(-2+x)^2]}² = {√[2+(2+x)^2]}²
[10+(-2+x)^2] = [2+(2+x)^2]
10 + 4 - 4x + x² = 2 + 4 + 4x + x²
-4x - 4x = -14 + 6
-8x = -8
x = -8/-8
x = 1
P = (1,0,0)
*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
08/04/2016
Sepauto - SSRC
*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
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