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what is the equation of the line that passes through (-9,12) and is perpendicular to the line whose equation is y=1/3x+6

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Apsweetangel9id
d: 1/3x-y+6, A(-9, 12)
md=-b/a=1/1/3=3, md=slope d

d`: y-yA=md(x-xA)
     y-12=3(x+9)
     y-12=3x+27
=> d`: 3x-y+15=0
Belcatyaid
y=kx+b
line is perpendicular y=1/3x+6⇒ k=-[tex] \frac{1}{ \frac{1}{3} } [/tex]=-3

 so line y=-3x+a
 this line that passes through (-9,12)⇔x=-9, y=12, so 12=-3(-9)+a
12=27+a
a=12-27
a=-15
y=-3x-15

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