A line segment is bisected by a line with the equation # 4 y - 6 x = 8 #. If one end of the line segment is at #( 7 , 3 )#, where is the other end?

Apology:
Even omitting some details, this explanation is quite long.

Consider the vertical line #x=7# which passes through the given point #(7,3)#
This vertical line will intersect #color(red)(4y-6x=8)# at #(7,25/2)#, a point which is #25/2-3=19/2# above the given point #color(green)(""(7,3))#
A point twice as far above #color(green)(""(7,3))# would be at #(7,3+2xx19/2)=(7,22)#
That is the line segment from #color(green)(""(7,3))# to #(7,22)# is bisected by #color(red)("4y-6x=8)#

Furthermore, as we can see from similar triangles any line parallel to #color(red)(4y-6x=8)# and through #(7,22)# gives an infinite collection of points, any one of which would serve as an endpoint with #color(green)(""(7,3))# for a line segment bisected by #color(red)(4y-6x=8)#

#color(red)(4y-6x=8)# has a slope of #3/2#
so any line parallel to it must also have a slope of #3/2#
and
if such a line passes through #(7,22)# then we can write its equation using the slope-point form:
#color(white)("XXX")y-22=3/2(x-7)#
or simplified as
#color(white)("XXX")color(purple)(2y-3x=23)#

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It is possible, since this question was asked under the heading "Perpendicular Bisectors" that it was intended that #color(red)(4y-6x=8)# should be the perpendicular bisector of a derived line segment.

In this case the perpendicular line to #color(red)(4y-6x=8)# passing through #color(green)(""(7,3))#
would have a slope of #-2/3# (the negative inverse of #color(red)(4y-6x=8)# and (again, working through the slope-point form)
an equation of #color(brown)(3y+2x=23)#

The system of equations:
#color(purple)(2y-3x=23)#
#color(brown)(3y+2x=23)#
can be solved for the point of intersection: #color(brown)(""(-23/15,46/5))#