A line segment is bisected by a line with the equation `-6 y + 3 x = 2`. If one end of the line segment is at `( 5 , 1 )`, where is the other end?

`color(red)(-6y+3x=2)` can be rewritten as `color(red)(y=1/2x-1/3)`

When `x=5`
`color(white)("XXX")y=1/2x-1/3color(white)("XX")rarrcolor(white)("XX")y=color(red)(13/6)`

The vertical distance from `color(green)(""(5,1))` to the point `color(red)(""(5,13/6))` on the line `color(red)(-6y+3x=2)` is
`color(white)("XXX")color(blue)(13/6-1=7/6)`

That is moving vertically the line `color(red)(-6y+3x=2)` is `color(blue)(7/6)` above the point `color(green)(""(5,1))`.

If we continue moving upward another `color(blue)(7/6)` units (for a total of `14/6=7/3` units above `color(green)(""(5,1))`)
we will reach the point `""(5,10/3))`
which will be the same distance above `color(red)(-6y+3x=2)` as `color(green)(""(5,1))` is below it.

That is `color(red)(-6y+3x=2)` bisects the line segment joining `color(green)(""(5,1))` and `color(purple)(""(5,10/3))`

In fact the line segment between `color(green)(""(5,1))` and any point on the line parallel to `color(red)(-6y+3x=2)` through `color(purple)(""(5,10/3))` will be bisected by `color(red)(-6y+3x=2)`

`color(red)(-6y+3x=2)` has a slope of `color(red)(1/2)`
so the line through `color(purple)(""(5,10/3))` will also have a slope of `color(red)(1/2)`
and using the slope-point form, it will have an equation of
`color(white)("XXX")y-color(purple)(10/3)=color(red)(1/2)(x-color(purple)(5))`
which can be simplified as
`color(white)("XXX")y=x/2+5/6`