Question

A line passes through `(6 ,2 )` and `(2 ,1 )`. A second line passes through `(3 ,2 )`. What is one other point that the second line may pass through if it is parallel to the first line?

Answer 1

`(1,3/2)` or graph{y = x/4 + 1/2 [-10, 10, -5, 5]} any point on the line `y = x/4 + 5/4`

Explanation:

First, find the gradient of the first line from the two points given.

gradient `=(rise)/(run)`

gradient `=(y1-y2)/(x1-x2)`

`(6,2) = (x1,y1)` and `(2,1) = (x2,y2)`

gradient `=(2-1)/(6-2)`

gradient `=(1)/(4)`
This is the first line: (`y = x/4 + 1/2`) - the equation was found by substituting a coordinate into `y = x/4 + c` to find c. (but you don't need to find the equation, this graph is just for explanation purposes)
graph{y = x/4 + 1/2 [-10, 10, -5, 5]}

The second line is parallel, meaning it has the same gradient, which is `1/4`. We also have one point `(3,2)`.

Substitute the point `(3,2)` and gradient of `1/4`in to `y = mx + c` where `m=`gradient to find the value of c.

`2 = 1/4(3) +c`

`c =5/4`

Therefore the equation for the second line is `y = x/4 + 5/4`, as shown:
graph{y = x/4 + 5/4 [-10, 10, -5, 5]}

Now, just choose any random x-value and find the corresponding y-value to get a point on that line.