Question

A line passes through `(9 ,2 )` and `( 3, 5 )`. A second line passes through `( 4, 1 )`. What is one other point that the second line may pass through if it is parallel to the first line?

Answer 1

`(0,3)`

Explanation:

For a second line to be parallel to the first, it must have the same slope, but not be the same line. In slope-intercept form, this means that the `b` portions of each `y=mx+b` equation must differ.

As we might recall, `m=(y_2-y_1)/(x_2-x_1)` Thus, for our first line:

`m_1 = (5-2)/(3-9) = 3/(-6) = -1/2`

We have `y_1 = m_1x_1+b`. Substituting (9, 2) for the x and y respectively...

`-> 2 = -1/2 (9)+b = -9/2 +b -> b = 13/2`

Giving us for the first line:

`y = -1/2 x + 13/2`.

NOTE: For the next part, we will be using prime notation, e.g. y', m'. These are not being used to denote Calculus derivatives, but rather to distinguish them from the original variable and slope.

For our second line, we must have the same slope, but a different intercept. We can find our Y intercept like we did above:

`y' = m'x'+b' -> 1 = -1/2 (4) + b = -2 + b -> b = 3 -> y' = -1/2 x + 3`

We know the y intercept is the value the function takes on when `x=0`. Thus, in this case:

`y'(0) = -1/2 (0) + 3 = 3`

Thus, one point this new parallel line could pass through is `(0,3)`.