Question

A line segment is bisected by a line with the equation `4 y - 3 x = 2`. If one end of the line segment is at `( 7 , 5 )`, where is the other end?

Answer 1

The other point is on the line `color(green)(4y-3x=5)`
(any point on this line will satisfy the given requirement)

Explanation:

Given
`color(white)("XXX")`The line `color(red)(4y-3x=2)`
and a point `color(purple)(""(7,5))`

Consider the vertical line through `color(purple)(""(7,5))`
This vertical line will intersect `color(red)(4y-3x=2)` at `color(red)(""(x=color(purple)(7),y))`
where `color(red)(y)` can be determined by solving
`color(white)("XXX")color(red)(4y-3xxcolor(purple)(7)=2)`

`color(white)("XXX")rarr color(red)(y=23/4)`

The distance from `color(purple)(""(7,5))` to `color(red)(""(7,23/4))`
is `color(red)(23/4)-color(purple)(5)=3/4`

So a point `color(green)(""(7,5+2xx3/4) = (7,13/2))` will be twice as far away from `color(purple)(""(7,5))` as the point `color(red)(""(7,23/4))` on the same vertical line

Note that any point on a line parallel to `color(red)(4y-3x=2` through `color(green)(""(7,13/2))` will also be twice as far away from `color(purple)(""(7,5))` as the point from `color(purple)(""(7,5))` on `color(red)(4y-3x=2)` to that point.

If `color(red)("L1")` parallel to `color(green)("L2")`
then `triangle ABC ~=triangle ADE`
`rarr abs(AB):abs(AD)=abs(AC):abs(AE)`

Since `color(red)(4y-3x=2)` has a slope of `color(red)(3/4)`

Our required line will also have a slope of `color(green)(3/4)`
and since it passes through `(""(7,13/2))`
using the slope-point form, we have
`color(white)("XXX")color(green)(y-13/2=3/4(x-7))`
or
`color(white)("XXX")color(green)4y-3x=5`

Answer 2

The other end-pt. lies on the line given by the eqn. `4y-3x=5.`

Explanation:

Suppose that the other end-pt. is `P(X,Y)`.

Let the given end-pt. be `Q(7,5),` and the given line be `L : 4y-3x=2.`

If `M` is the mid-pt. of the segment `PQ`, then co-ords. of `M` as obtained by using Section Formula for Mid-pt. are `=M((7+X)/2,(5+Y)/2)`

Now, `PQ` is bisected by `L` at `M`, so, `M in L.`

Therefore, co-ords. of `M` must satisfy the eqn. of `L.`

Hence, `4{(5+Y)/2}-3{(7+X)/2}=2 rArr 20+4Y-21-3X=4,` i.e., `4Y-3X=5,` as Sir Alan P. has readily derived!

This shows that :
(i) the co-ords. of other end-pt. can not be uniquely derived under the given conds.
(ii) What we can say about it (the other end-pt.) is that it lies on `: 4y-3x=5.` This eqn. represents a line `||` to `L`.