Question

A line segment is bisected by a line with the equation `-7 y + 3 x = 2`. If one end of the line segment is at `( 2 , 4 )`, where is the other end?

Answer 1

Infinite solutions: any of the points in the line `y=(3/7)x-26/7`, including the point `(110/29,-424/203)` when the segment is perpendicularly bisected.

Explanation:

First we should notice that the point (2,4) isn't in the line `3x-7y-2=0` since:
`3*2-7*4-2=6-28-2=-24`

But as the problem is stated it admits infinite solutions.

Calling AB the line segment with M the midpoint we have:
`A(2,4)`
`M((2+x_B)/2,(4+y_B)/2)`
`B(x_B,y_B)`

The orthogonal projection of A onto the line is also M when the segment is perpendicular bisected.
The slope of the line `y=(3x-2)/7`is
`k=3/7`
So the slope of line perpendicular to the aforementioned is
`p=-1/k=-7/3`
And the equation of such a perpendicular line passing through A is
`y-4=-7/3(x-2)` => `y=(-7x+14)/3+4` => `y=(-7x+26)/3`

Combining the equations of the 2 lines
`(3x-2)/7=(-7x+26)/3` => `9x-6=-49+162` => `58x=168` => `x=84/29`
`-> y=(252/29-2)/7` => `y=194/203`
And `M(84/29,194/203)`

Finding B
`x_M=1+x_B/2` => `84/29=1+x_B/2`=>`x_B=55/29*2=110/29`
`y_M=2+y_B/2` =>`194/203=2+y_B/2` =>`y_B=-212/203*2=-424/203`
So when the segment AB is perpendicularly bisected by the line mentioned in the problem
`B(110/29,-424/203)`

But since it is not required that the bisection happens perpendicularly the set of possible points B lays in a line parallel to the bisecting line and passing through `(110/29,-424/203)`, itself one of the possible locations of B. Then the locus in which the point B can be located is
`y+424/203=(3/7)(x-110/29)`
`y=(3/7)x-330/203-424/203`
`y=(3/7)x-26/7`