SOLUTION: Hey, I've been looking all over the internet for a way to convert (y=(m)x+b) into point-slope form. I can't remember though whether this only applies to quadratic equations. If you

Question 527124: Hey, I've been looking all over the internet for a way to convert (y=(m)x+b) into point-slope form. I can't remember though whether this only applies to quadratic equations. If you consider this two questions, my apologies, I'm new to this.
Thanks.
Found 2 solutions by jim_thompson5910, scott8148:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Point slope form is

y - y1 = m(x - x1)

where (x1, y1) is a point that lies on the graph and m is the slope

From y = mx+b, the slope will be given. It'll be up to you find a point that lies on the graph. The easiest point is the y-intercept, which is (0, b). So x1 = 0 and y1 = b

This means that

y = mx+b

converts to

y - b = m(x - 0)

where y1 = b

If you need more help, email me at jim_thompson5910@hotmail.com

Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you

Jim
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
for the line through the point (d,e) ___ y - e = m(x - d)

since b is (0,b); e = b + md

not sure how useful this is

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