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#
