How do you write an equation of a line passing through (-3, 4), perpendicular to #3y=x-2#?

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Answer 1 · 2018-05-03T01:56:35

#3x+y+5=0# is the required equation of the straight line. graph{(3x+y+5)(x-3y-2)=0 [-8.44, 2.66, -4.17, 1.38]}

Any line perpendicular to #ax+by+c=0# is #bx-ay+k=0# where k is constant.

Given equation is

#rarr3y=x-2#

#rarrx-3y=2#

Any line perpendicular to #x-3y=2# will be #3x+y+k=0#

As #3x+y+k=0# passes through #(-3,4)#, we have,

#rarr3*(-3)+4+k=0#

#rarr-9+4+k=0#

#rarrk=5#

So, the required equation of the straight line is #3x+y+5=0#