Question

A line passes through `(8 ,2 )` and `(6 ,7 )`. A second line passes through `(3 ,4 )`. What is one other point that the second line may pass through if it is parallel to the first line?

Answer 1

Any point on the line `y=(23-5x)/2` is acceptable. For example `(1, 9)` is a correct answer.

Explanation:

To determine the slope, `m` of a line we can use the formula

`m=(y_2-y_1)/(x_2-x_1)`

where `(x_1, y_1)` and `(x_2, y_2)` are any two distinct points on the line. Here, the two points for the first line are `(8, 2)` and `(6, 7)` so the slope of the line is

`m=(7-2)/(6-8)=-5/2`

The second line must also abide by this slope definition, yet we only know one point on the line. Let's call the coordinates of the second point `(x, y)`. The slope formula for the second line then gives

`m=-5/2=(y-4)/(x-3)`

Let's solve this equation for y by multiplying both sides of this equation by `(x-3)` and then adding 4 to both sides of this equation.

`y=-5/2(x-3)+4=-5/2x+15/2+4=-5/2x+23/2`

So our line has a slope of -5/2 and a `y`-intercept of `23/2`.

The problem statement asks for ANY other point on this line. This means we can pick any `x`-value we want (except for 3) and then calculate the value of y. I'm going to choose `x=1`. If `x=1`, then

`y=-5/2(1)+23/2=18/2=9`

So `(1, 9)` is a point on the second line.