Let's do a parametric solution, which I think is slightly less work.

Let's write the given line

#-7x + 3y = 2 quad quad quad quad quad quad quad quad y = 7/3 x + 2/3#

I write it this way with #x# first so I don't accidentally substitute in a #y# value for an #x# value. The line has a slope of #7/3# so a direction vector of #(3,7)# (for every increase in #x# by #3# we see #y# increase by #7#). This means the direction vector of the perpendicular is #(7,-3).#

The perpendicular through #(7,3)# is thus

#(x,y)=(7,3) + t(7,-3)=(7+7t,3-3t)#.

This meets the original line when

#-7(7 + 7t) + 3(3-3t) = 2 #

#-58t=42#

# t = -42/58=-21/29 #

When #t=0# we're at #(7,3),# one end of the segment, and when #t=-21/29# we're at the bisection point. So we double and get #t=-42/29# gives the other end of the segment:

#(x,y)=(7,3)+ (-42/29)(7,-3) = (-91/29, 213/29)#

That's our answer.

Check:

We check bisector then we check perpendicular.

The midpoint of the segment is

# ( (7 + -91/29)/2 , ( 3+ 213/29 )/2 ) = (56/29, 150/29)#

We check that's on #-7x + 3y = 2 #

# - 7 (56/29) + 3 ( 150/29) = 2 quad sqrt #

Let's check it's a zero dot product of the difference of the segment endpoints with the direction vector #(3,7)#:

# 3 (-91/29 - 7) + 7( 213/29 - 3) = 0 quad sqrt#