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

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`