Posted by Anonymous on Friday, December 6, 2013 at 10:56am.
I have an assignment to write an equation to point-slope form, then to standard form, and then to slope-intercept form. I have used the points (-2,-1) or (1,2), but I used (-2,-1).
Now, I have the point-slope form of y + 1= 1(x + 2) and the standard form of y – 1x = -1. When I convert into slope-intercept form, I am getting the slope as -1, which is the problem. Could someone please tell me where I am going wrong?
Algebra - Anonymous, Friday, December 6, 2013 at 11:17am
Actually, the slope is not -1, but the y-intercept comes up as -1.
Algebra - Reiny, Friday, December 6, 2013 at 12:40pm
your slope is correct as +1
your equation of
y+1 = 1(x+2) is also correct
then y+1 = x+2
y = x+1 ---> slope-yintercept form
y-x = 1
x - y = -1 ----> standard form ( I multiplied each term by -1)
Algebra - Anonymous, Friday, December 6, 2013 at 12:51pm
So, how did you multiply the terms by -1?
Algebra - Anonymous, Friday, December 6, 2013 at 1:17pm
And how do you rewrite x - y = -1 into slope-intercept form?
Algebra - Reiny, Friday, December 6, 2013 at 4:32pm
in standard form we usually lead with the x term with a positive coefficient, so I wanted a +x as my first term, it was -x .
to change a standard form to slope- yintercept form usually takes 2 steps:
1. move all terms except the y term to the right
4x - 5y = 8 ---≥ -5y = -4x + 8
2. divide each term by the coefficient of the y term
-5y = -4x + 8 ----> y = (-4/-5)x + 8/-5
y = (4/5)x - 8/5 , just an example, has nothing to do with your question
apply those 2 steps to your question.
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