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Sagot :
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
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
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
P.s.picture to test solutions in the application
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
P.s.picture to test solutions in the application

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