Connecting...

This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Algebra 1
For more information, please see full course syllabus of Algebra 1
Algebra 1 Point Slope Form of an Equation
Lecture Description
Dr. Eaton covers the Point Slope Form of an Equation by first manipulating this new form into other more familiar equations before covering the representation of slope and a test point. This lesson ends with additional examples.
Bookmark & Share
Embed
Share this knowledge with your friends!
Copy & Paste this embed code into your website’s HTML
Please ensure that your website editor is in text mode when you paste the code.(In Wordpress, the mode button is on the top right corner.)
×
Since this lesson is not free, only the preview will appear on your website.
- - Allow users to view the embedded video in full-size.
Next Lecture
Previous Lecture
0 answers
Post by Kapil Patel on February 14, 2017
using the given conditions to write an equation for the line in point slope form and slope-intercept form passing through (-4,13)and (1,-2).
1 answer
Tue Jun 18, 2013 6:38 PM
Post by Taylor Wright on June 18, 2013
Wouldn't it be easier to just distribute the 2/3 to the (x+4) and then add 3 to both sides?
1 answer
Fri May 17, 2013 7:18 PM
Post by Erika Porter on May 17, 2013
Dr. Eaton,
On example 3, you multiplied y-3=2/3(x+4)by 3 giving you:
3(y-3)=[2/3(x+4)]3
which then gave you 3y-9=2(x+4).
My question is: Why was the (x+4) not multiplied by 3 as well?
However, when I worked the problem, I substituted the known values for x (-4),y(3), and m(2/3) into the slope intercept form to solve for b.
3=(2/3)(-4)+b
solving for b...b=17/3
then...putting into slope intercept form I came up with y=2/3x+17/3 which is the same answer as you.
I am perplexed.
thanks,
Jake
0 answers
Post by S A A Mazeed Sumon on April 29, 2010
Did Not Close The First Bracket for X....