A line segment is bisected by a line with the equation `9 y - 2 x = 5`. If one end of the line segment is at `( 7 , 3 )`, where is the other end?

If the line is the perpendicular bisector of the segment, the other endpoint is uniquely determined.

If the line is just a bisector, as asked, each point on the line is the bisector of some segment whose endpoint is `(7,3)` and whose other endpoint sweeps out a curve. Let's see if we can figure out its equation.

If we're told `(p,q)` is the midpoint of a segment with endpoint `(a,b)=(7,3)` the other endpoint `(x,y)` satisfies:

`(x,y)-(p,q)=(p,q)-(a,b)`

`(x,y)=(2p-a,2q-b)`

`(x,y)=(2p-7,2q-3)`

We have `(p,q)` on the line so `2p=9q-5`.

`(x,y)=(9q-12,2q-3)=(-12,-3)+q(9,2)`

The other endpoint `(x,y)` makes a line through `(-12,-3)` with direction vector `(9,2)` meaning slope `2/9.` Nonparametrically, that's the line

`y+3 = 2/9(x+12)`

`9y+27=2x+24`

`9y-2x=-3`

That's parallel to the original line, as far from it as `(7,3)` is.

Now let's ask which `(x,y)` makes a perpendicular bisector to the line?

The direction of the segment is `(x-a,y-b)=(x-7,y-3).` The direction of the line `-2x+9y=5` is `(9,2)` meaning for every `9` units `x` increases we increase `y` by `2` units.

We have perpendicularity when the dot product is zero:

`(x-7,y-3) cdot(9,2) = 0`

`9(9q-12 - 7) + 2( 2q-3-3) = 0`

`85q - 183 = 0`

`q = 183/85`

`(x,y) = (9q-12,2q-3)=(627/85, 111/85)`

Check:

`( (627/85,111/85)-(7,3) ) cdot (9,2) = 0 quad sqrt`