# confused :(

• Jan 20th 2008, 04:59 PM
Morzilla
confused :(
Ok, I am graphing P=(1,2) with slope m=3. This is how I worked out

$y-2=3(x-1)$

= $y-2=3x-3$

= $y=3x-3+2$

= $y=3x-1$

If x=0 then;

$y=3(0)-1$

= $y=-1$

and so (0,-1)

this would be my second point no, or is this my run? In checking with the book's answer to see if I had done this right it showed my second point to be (2,5) with a run of 1 and a rise of 3......Did I do something wrong in the calculations!?

Thanks!!
• Jan 20th 2008, 10:04 PM
You are right. When graphing a line by getting 2 points and drawing a line through them you can use any 2 points on a line. The book simply chose its points differently.
• Jan 21st 2008, 08:54 PM
Morzilla
Quote:

You are right. When graphing a line by getting 2 points and drawing a line through them you can use any 2 points on a line. The book simply chose its points differently.

.....oh, ok. So in this case, having my second points (0,-1) I can then find the run of the slope no?!

Thanks !!!

(Handshake)
• Jan 23rd 2008, 07:25 PM
They are pretty much only useful to find the slope using $m = \frac {rise}{run}$