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

y=kx+b line is perpendicular y=1/3x+6⇒ k=-[latex] \frac{1}{ \frac{1}{3} } [/latex]=-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