What is the equation in standard form of a perpendicular line tht passes through (5,1)?

all perpendicular lines have the NEGATIVE RECIPICAL
so it should be something like-
y= -1/3x+0
but you y-intercept(the last number, in this case 0) could be otherwise.

There is a linear equation in point-slope form:
y - 3 = 3(x + 1),
which tells us that a line g with slope m=3 passes through P1(-1, 3).
g := y = 3x + 6
And you want the equation for the perpendicular line g' that passes through P2(5, 1).

The slope m' of g' is:
m' = -1 m^-1 . Thus, for m = 3, m' = -1/3.

Now we have a point P2(5, 1) and a slope m'=-1/3, so we can define the line g'

g' := y - 1 = -1/3(x - 5)
y = -x/3 + 5/3 + 1
y = -x/3 + 8/3 in standard form

Btw, the intersection point of g and g' is P1:

3x + 6 = -x/3 + 8/3
9x + 18 = -x + 8
10x = -10
x = -1
Insert that in either equation (g or g')
y = 3.