Question 935263
{{{Ax + By = C}}}, {{{A >= 0}}}.

({{{0}}},{{{-4}}}); perpendicular to {{{x + 3y = 9}}}

 {{{3y =-x+ 9}}}

{{{y =-(1/3)x+ 3}}}


*[invoke equation_parallel_or_perpendicular "perpendicular", "-1/3", 3, 0, -4]


{{{y =3x-4}}}in {{{Ax + By = C}}} form: {{{4=3x-y}}}

or
{{{3x-y =4}}} where {{{A= 3}}}=>{{{A >=0}}}