Obter a reta que passa por P(3,-1) e é paralela a r: 3x+14y -17=0
obrigado!
3x+14y -17=0
14y = -3x +17........(:14)
y = (-3/14)x + 17/14...........=> a = -3/14
y-yp = a.(x-xp)
y+1= -3(x-3)/14
-3x+9 = 14(y+1)
-3x+9 -14y -14 =0
-3x -14y -5 = 0..........*(-1)
3x +14y +5 = 0
(y+1)=-3/14(x-3)
14(y+1)=-3x+9
14y+14=-3x+9
3x+14y+5=0
3x+14y + Q=0
falta achar o Q
3*3+14*(-1) + Q = 0
9-14 +Q =0
Q= 5
3x+14y+5 =0 <------ respostq
=================================
3x + 14y - 17 = 0
14y = -3x + 17
y = -3x/14 + 17/14
reta //
y = -3x/14 + b
-1 = -3*3/14 + b
b = -1 + 9/14 = -5/14
y = -3x/14 - 5/14
equação
-3x - 14y - 5 = 0
pronto
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Verified answer
3x+14y -17=0
14y = -3x +17........(:14)
y = (-3/14)x + 17/14...........=> a = -3/14
y-yp = a.(x-xp)
y+1= -3(x-3)/14
-3x+9 = 14(y+1)
-3x+9 -14y -14 =0
-3x -14y -5 = 0..........*(-1)
3x +14y +5 = 0
(y+1)=-3/14(x-3)
14(y+1)=-3x+9
14y+14=-3x+9
3x+14y+5=0
3x+14y + Q=0
falta achar o Q
3*3+14*(-1) + Q = 0
9-14 +Q =0
Q= 5
3x+14y+5 =0 <------ respostq
=================================
3x + 14y - 17 = 0
14y = -3x + 17
y = -3x/14 + 17/14
reta //
y = -3x/14 + b
-1 = -3*3/14 + b
b = -1 + 9/14 = -5/14
y = -3x/14 - 5/14
equação
-3x - 14y - 5 = 0
pronto