zairac27
zairac27 zairac27
  • 02-10-2016
  • Mathematics
contestada

(-3,-3) ; parallel to -x+3y=9

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Ckaranja
Ckaranja Ckaranja
  • 11-10-2016
Rewriting the equation in the form y=mx+c
-x+3y=9
3y=x+9
y=[tex] \frac{1}{3} x+3[/tex]
Gradient is 1/3
Since parallel lines have similar gradients,
[tex] \frac{y--3}{x--3}= \frac{1}{3} [/tex]
y+3=[tex] \frac{1}{3} (x+3)[/tex]
y+3=[tex] \frac{1x}{3} +1[/tex]
y=[tex] \frac{1x}{3} -2[/tex]
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