Question

A line passes through `(4 ,7 )` and `(2 ,8 )`. A second line passes through `(1 ,5 )`. What is one other point that the second line may pass through if it is parallel to the first line?

Answer 1

The point `(0, 11/2)`.

Explanation:

The equation of a line is `y=mx+q`
The line that passes from `(4,7)` and `(2,8)` can be obtained substituting the coordinates as `x` and `y` and doing the system

`7=4m+q`
`8=2m+q`

We can subtract the second from the first to remove `q` obtaining

`7-8=4m-2m+\cancel(q)-\cancel(q)`

`-1=2m` and `m=-1/2`.

We can also calculate `q`, even if it is not relevant for our problem, using one of the two equations

`7=-1/2*4+q`

`7=-2+q`

`q=9`

The equation of the first line is then

`y=-1/2x+9`

The second line is parallel, it means that the slope is the same `-1/2`.
So the equation of the parallel is

`y=-1/2x+r` where `r` is the intercept of the second line. To find it we know that the line passes from `(1,5)`, then

`5=-1/2*1+r`

`5+1/2=r`

`11/2=r`

The equation of the parallel is

`y=-1/2x+11/2`.

To find another point in this line it is enough to give a value to `x` and calculate the `y`.
For example `x=0`, `y=-1/2*0+11/2=11/2`, then the point `(0, 11/2)` is another point of the parallel.