# Thread: Proof of y-intercept

1. ## Proof of y-intercept

Hi all, I'm working through some problems at the moment and I am a little stuck on an algebraic rearrangement of a coordinate geometry proof. The problem is as follows:

We have a line passing through two points A: (x1, y1) and B: (x2, y2) (where x1 doesn't equal x2). Let P: (x, y) be any point on the line. We can define the equation of the line as:

[1] ((y-y1)/(x-x1))=((y2-y1)/x2-x1))

Putting this into the form y = mx +c gives

[2] m= ((y2-y1)/(x2-x1)) ... I get this, but then it says that

[3] c= (((x2y1)-(x1y2))/(x2-x1)) ... I don't understand how we arrive at this expression from rearranging [1], could someone please help?

Thanks,

2. ## Re: Proof of y-intercept

It's pretty much just algebra- specifically the "distributive law".

You have $\frac{y- y_1}{x- x_1}= \frac{y_2- y_1}{x_2- x_1}$.

Multiply both sides by $x- x_1$: $y- y_1= \frac{y_2- y_1}{x_2- x_1}\left(x- x_1\right)$

Now "distribute" the fraction on the right over the " $x$" and " $x_1$"

$y- y_1= \frac{y_2- y1}{x_2- x_1}x- \frac{y_2- y_1}{x_2- x_1}x_1$

That first part, on the right, $\frac{y_2- y_1}{x_2- x_1}x$ is the "mx" which you say you understand.

Add $y_1$ to both sides
$y= \frac{y_2- y1}{x_2- x_1}x- \frac{y_2- y_1}{x_2- x_1}x_1+ y_1$

and add $y_1$ to that last fraction. That part alone, c, will be
$-\frac{y_2- y_1}{x_2- x_1}x_1+ y_1= \frac{-x_1(y_2- y_1)}{x_2- x_1}+ \frac{y_1(x_2- x_1}{x_2- x_1}$
$= \frac{-x_1y_2+ x_1y_1+ x_2y_1- x_1y_1}{x_2- x_1}$

The " $x_1y_1$" terms cancel leaving
$\frac{x_2y_1- x_1y_2}{x_2- x_1}$

3. ## Re: Proof of y-intercept

That's brilliant, thanks for such a quick response. I was getting stuck at the "distribute" part. Thanks again