m=5/2
(y+3)=5/2(x-2)
2(y+3)=5x-10
2y+6=5x-10
5x-2y-16=0
Vamos achar o coeficiente angular da segunda equação:
-2y=-5x-1
2y=5x+1
y=(5x+1)/2
y=(5x)/2+(1/2)
Coeficiente angular = 5/2
Se é paralela ==> mr=ms
y-yo=m.(x-xo)
y+3=(5/2).(x-2)
y+3=5/2x-10/2
y=5/2x-5
y=5/2x-5-3
y=5/2x-8 (reduzida)
y-5/2x+8=0
2y/2-5x/2+16/2=0/2
2y-5x+16=0 . (-1)
-2y+5x-16=0
5x-2y-16=0 (geral)
5x-2y=16
5x/16-2y/16=16/16
5x/16-2y/16=1
5x/16-y/4=1 (segmentária)
Ola Ana eliza
5x - 2y + k = 0
calcule k com o ponto P(2,-3) assim
5*2 - 2*(-3) + k = 0
10 + 6 + k = 0
k = -16
5x - 2y - 16 = 0
pronto
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Answers & Comments
m=5/2
(y+3)=5/2(x-2)
2(y+3)=5x-10
2y+6=5x-10
5x-2y-16=0
Vamos achar o coeficiente angular da segunda equação:
-2y=-5x-1
2y=5x+1
y=(5x+1)/2
y=(5x)/2+(1/2)
Coeficiente angular = 5/2
Se é paralela ==> mr=ms
y-yo=m.(x-xo)
y+3=(5/2).(x-2)
y+3=5/2x-10/2
y=5/2x-5
y=5/2x-5-3
y=5/2x-8 (reduzida)
y-5/2x+8=0
2y/2-5x/2+16/2=0/2
2y-5x+16=0 . (-1)
-2y+5x-16=0
5x-2y-16=0 (geral)
5x-2y=16
5x/16-2y/16=16/16
5x/16-2y/16=1
5x/16-y/4=1 (segmentária)
Ola Ana eliza
5x - 2y + k = 0
calcule k com o ponto P(2,-3) assim
5*2 - 2*(-3) + k = 0
10 + 6 + k = 0
k = -16
5x - 2y - 16 = 0
pronto