Question

A line segment is bisected by a line with the equation `5 y + 2 x = 1`. If one end of the line segment is at `(3 ,4 )`, where is the other end?

Answer 1

the other is end is at `(-13/29, -134/29)`

Explanation:

The given line is `5y+2x=1`
The line perpendicular to this line and passing thru point `U(3, 4)` is
`y-4=(5/2)(x-3)` by the two-point form
or `5x-2y=7`
Simultaneous solution of
`2x+5y=1` and `5x-2y=7` yields point `I(37/29, -9/29)`

Let `v=`vertical distance from point `I(37/29, -9/29)` to `U(3, 4)`
`v=4--9/29=125/29`
Let `h=`horizontal distance from point `I(37/29, -9/29)` to `U(3, 4)`
`h=3-37/29=50/29`
Let the other end point be `D(x_o, y_o)`

`x_o=37/29-h=37/29-50/29= -13/29`
`y_o=-9/29-125/29=-134/29`

Check by distance formula from point `I(37/29, -9/29)` to `U(3, 4)`
`d_1=sqrt((3-37/29)^2+(4--9/29)^2)=(25sqrt(29))/29`

Check by distance formula from point `I(37/29, -9/29)` to `D(-13/29, -134/29)`
`d_2=sqrt((37/29--13/29)^2+(-9/29--134/29)^2)=(25sqrt(29))/29`

Therefore `d_1=d_2`

God bless....I hope the explanation is useful.