Question

How do I find the equation of the line which passes through the points `(8, 1)` and `(-2, 3)`?

Answer 1

There are several ways to find the equation of the line which passes through two points. The one I like the most is the vectorial method.

Let `v` be the vector that connects two points `P_0=(x_0,y_0)` and `P_1=(x_1,y_1)` and `k` a scalar, we can find any generic point `P=(x,y)` in their line by adding `k*v` to the coordinates of `P_0`.

`P = P_0 + k*v -> (x,y) = (x_0,y_0) + k*v`
`v = (x_1-x_0,y_1-y_0)`
`x = x_0 + k*(x_1-x_0) -> k = ((x-x_0))/((x_1-x_0))`
`y = y_0 + k*(y_1-y_0) -> k = ((y-y_0))/((y_1-y_0))`
`((x-x_0))/((x_1-x_0)) = ((y-y_0))/((y_1-y_0))`
`y = (x-x_0)*((y_1-y_0))/((x_1-x_0))`//

Hope it helps.

Or

1) Find the slope using ...

`m=(y_2-y_1)/(x_2-x_1)=(1-3)/(8-(-2))=(-2)/(8+2)=(-2)/10=-1/5`

2) Find the `y`-intercept, `b`, by using the slope intercept equation, `y=mx+b`

Use either of the points. I will use the `(8,1)`.

`1=(-1/5)(8)+b`

`1=(-8/5)+b`

`1+(8/5)=b`

`5/5+(8/5)=b`

`13/5=b`

3) Rewrite the slope intercept form, `y=mx+b`, and substitute in the `y`-intercept, `b`, and the slope, `m`.

`y=-1/5x+13/5`

In the image below see the function graphed. The green squares are the points (8,1) and (-2,3). Notice that they are on the line.