How do you write an equation in slope-intercept form for a line with (4, -1) and perpendicular to a line with equation 3x + y = 4?

A linear equation in the form #Ax+By=C# has a slope of #-A/B# [Note 1, below]

The given line #3x+y=4# has a slope of #-3/1=-3#

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If two lines are perpendicular the slope of one is the negative reciprocal of the other.

Any line perpendicular to #3x+y=4# has a slope of #color(green)(+1/3)#

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A line with a slope of #color(green)m# through a point #(color(red)(x_1),color(blue)(y_1))#
can be written in slope-point form as:
#color(white)("XXX")y-color(blue)(y_1)=color(green)m(x-color(red)(x_1))# [Note 2, below]

A line perpendicular to #3x+y=4# through the point #(color(red)4,color(blue)(-1))#
can be written as:
#color(white)("XXX")y-color(blue)((-1))=color(green)(1/3)(x-color(red)4)#
which would typically be simplified as
#color(white)("XXX")y+1=1/3(x-4)#

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Here is a graph with these two equations and the required point for verification:

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Note 1
If you are not familiar with slope of #Ax+By=C# being #-A/B#
it can be easily derived by converting to slope intercept form:
#color(white)("XXX")Ax+By=C#
#color(white)("XXX")rarr By=-Ax+C#
#color(white)("XXX")rarr y=-A/Bx+C/B# with slope of #-A/B# and y-intercept #C/B#

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Note 2
For some the more common general slope-point form might be
#color(white)("XXX")((y-color(blue)y_1))/(x-color(red)((x_1)))=color(green)m#
I prefer the form that I have used above since it avoids the exclusion problem when #x=color(red)(x_1)#