Question

A line passes through `(5 ,1 )` and `(6 ,5 )`. A second line passes through `(4 ,3 )`. What is one other point that the second line may pass through if it is parallel to the first line?

Answer 1

`(1,-9)`

Explanation:

So the steps are find the gradient of the first line (it will be the same gradient in the second line as it is parallel) then write the second line in the from

`y = ax + b`

(where `a` is the gradient we found) and solve for b given the point `(4,3)`. Then find any point by substituting values.

Step 1. Gradient

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

`m = (1 - 5) / (5 - 6) ->` (from the first two points)

`"gradient" = 4`

Step 2. Equation

Second line's equation is `y = 4x + b`
Solve for `b` using `(4,3)`

`3 = 4 (4) + b`

`b = 3-16`

`b= -13`

Hence the equation is

`y = 4x -13`

Step 3. Find point

Next we find a point. Let `x = 1` (`x` can be any value, then just substitute it in to find `y`)

`y = 4 (1) - 13`

`y = -9`

so the point is `(1, -9)`