A line segment is bisected by a line with the equation `- 6 y + 2 x = 3`. If one end of the line segment is at `( 4 , 8 )`, where is the other end?

Consider a horizontal line segment from `(4,8)` bisected by `-6y+2x=3`
The equation of the horizontal line segment from `(4,8)` is
`color(white)("XXX")y=8`

Noting that `-6y+2x=3 rArr x=(6y+3)/2`

The intersection of this horizontal line segment with the given bisector line will occur at
`color(white)("XXX")((6(8)+3)/2,8)=(51/2,8)`

Continuing to travel horizontally a point twice as far away from `(4,8)` as `(51/2,8)`
will be at `(51,8)`

That is `(51,8)` is one possible endpoint for a line segment form `(4,8)` bisected by `-6y+2x=3` would be `(51,8)`

For any point which could be such an endpoint, a line through this point parallel to the bisecting line will provide all possible bisected line segment endpoints.

`-6y+2x=3` (and all lines parallel to it) has a slope of `1/3`

Using the previously determined possible endpoint `(51,8)` and this slope of `1/3`
we can write the slope-point version:
`color(white)("XXX")y-8=1/3(x-51)`

`color(white)("XXX")3y-24=x-51`

`color(white)("XXX")3y-x=-27`

Any solution to the equation `3y-x=-27` will provide a valid endpoint for a line segment from `(4,8)` which will be bisected by `6y-2x=3`

Here is a graph of the point and the two lines in question:
graph{(-6y+2x-3)(3y-x+27)((x-4)^2+(y-8)^2-0.02)=0 [-25.65, 25.64, -12.83, 12.81]}