Question 133144
First convert the standard equation {{{x+3y=6}}} into slope intercept form


*[invoke converting_linear_equations "standard_to_slope-intercept", 1, 3, 6, 2, 1]




Now let's find the equation of the line that is perpendicular to {{{y=(-1/3)x+2}}} which goes through (-3,5)


*[invoke equation_parallel_or_perpendicular "perpendicular", "-1/3", "2", -3,5]


Now let's convert {{{y=3x+14}}} to standard form



{{{y-3x=14}}} Subtract 3x from both sides



{{{-3x+y=14}}} Rearrange the terms




So the equation that is perpendicular to {{{x+3y=6}}} and goes through (-3,5) is


{{{-3x+y=14}}}