How do you write an equation Perpendicular to the line -3x + y = -7 and Contains the point of: (3,1)?

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Answer 1 · 2016-07-22T04:13:45

#x+3y-6=0#.

Let #l : -3x+y=-7#, i.e., #y=3x-7#, be the given line, #l'# the

reqd. line, and the given pt. #A(3,1) in l'#.

Observe that the slope of #l=3#, &, as #l' bot l,#the slope of l' has to

be #-1/3#.

Further, #A(3,1) in l'#.

Hence, using the Slope-Point Form for #l'#, we find,

eqn. of # l' : y-1=-1/3(x-3), i.e., 3y-3=-x+3, or, x+3y-6=0#.