How do you find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5)?

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Answer 1 · 2017-02-01T04:57:54

Desired equation is #3x-y+14=0#

Equation of a line that is perpendicular to #Ax+By=C#

is of the form #Bx-Ay=k# i.e. reversing te coefficients of #x# and #y# and changing sign of one of them.

Hence, the line perpendicular to #x+3y=6# has equation of the form

#3x-y=k# and as it passes through #(-3,5)#

we should have #3xx(-3)-5=k#

and therefore #k=-9-5=-14#

and desired equation is #3x-y=-14# or #3x-y+14=0#
graph{(3x-y+14)(x+3y-6)=0 [-11.38, 8.62, -3.12, 6.88]}