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)(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)(p,q) is the midpoint of a segment with endpoint (a,b)=(7,3)(a,b)=(7,3) the other endpoint (x,y)(x,y) satisfies:
(x,y)-(p,q)=(p,q)-(a,b)(x,y)−(p,q)=(p,q)−(a,b)
(x,y)=(2p-a,2q-b)(x,y)=(2p−a,2q−b)
(x,y)=(2p-7,2q-3)(x,y)=(2p−7,2q−3)
We have (p,q)(p,q) on the line so 2p=9q-52p=9q−5.
(x,y)=(9q-12,2q-3)=(-12,-3)+q(9,2)(x,y)=(9q−12,2q−3)=(−12,−3)+q(9,2)
The other endpoint (x,y)(x,y) makes a line through (-12,-3)(−12,−3) with direction vector (9,2)(9,2) meaning slope 2/9.29. Nonparametrically, that's the line
y+3 = 2/9(x+12)y+3=29(x+12)
9y+27=2x+249y+27=2x+24
9y-2x=-39y−2x=−3
That's parallel to the original line, as far from it as (7,3)(7,3) is.
Now let's ask which (x,y)(x,y) makes a perpendicular bisector to the line?
The direction of the segment is (x-a,y-b)=(x-7,y-3).(x−a,y−b)=(x−7,y−3). The direction of the line -2x+9y=5−2x+9y=5 is (9,2)(9,2) meaning for every 99 units xx increases we increase yy by 22 units.
We have perpendicularity when the dot product is zero:
(x-7,y-3) cdot(9,2) = 0(x−7,y−3)⋅(9,2)=0
9(9q-12 - 7) + 2( 2q-3-3) = 09(9q−12−7)+2(2q−3−3)=0
85q - 183 = 085q−183=0
q = 183/85q=18385
(x,y) = (9q-12,2q-3)=(627/85, 111/85)(x,y)=(9q−12,2q−3)=(62785,11185)
Check:
( (627/85,111/85)-(7,3) ) cdot (9,2) = 0 quad sqrt
