Question

A line segment is bisected by a line with the equation `- y + 7 x = 1`. If one end of the line segment is at `(1 ,3 )`, where is the other end?

Answer 1

Any point on the line `-y+7x=-2`

Explanation:

Note that `(x,y)=(1,6)` is a point on the line -y+7x=1#

The distance from `(1,3)` to `(1,6)` is `3`

The distance from `(1,3)` to `(1,6+3)=(1,9)` is `6`

`(1,3), (1,6), and (1,9)` are co-linear.

Therefore the line segment from `(1,3)` to `(1,9)` is bisected by the line `-y+7x=1`

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Furthermore a line segment between `(1,3)` and any point on a line through `(1,9)` parallel to `-y+7x=1`
will also be bisected by `-y+7x=1`

The equation of this line is
`color(white)("XXX")y-9=7(x-1)`
or
`color(white)("XXX")y-7x=2`
or, in a form similar to the given equation
`color(white)("XXX")-y+7x=-2`

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See similar problem solution at:
https://socratic.org/questions/a-line-segment-is-bisected-by-a-line-with-the-equation-2-y-x-1-if-one-end-of-the#272239
for a more detailed solution with diagrams