# Coordinate geometry question

• Nov 13th 2007, 10:23 AM
hymnseeker
Coordinate geometry question
Just a really quick one, suppose I have a line of equation: y=1/2x+5/2

The next part of the question asks for this in the form of ax+by+c=0

How do I convert from y=mx+c to ax+by+c=0?

Thank youuu! :)
• Nov 13th 2007, 10:24 AM
Jhevon
Quote:

Originally Posted by hymnseeker
Just a really quick one, suppose I have a line of equation: y=1/2x+5/2

The next part of the question asks for this in the form of ax+by+c=0

How do I convert from y=mx+c to ax+by+c=0?

Thank youuu! :)

bring everything to one side and leave 0 on the other side and (in this case) multiply through by 2 when done
• Nov 13th 2007, 10:26 AM
hymnseeker
Thanks for the reply, so just let y=0?

1/2x+5/2=0

Does this satisfy what the question wants in asking for ax+by+c=0?
• Nov 13th 2007, 10:29 AM
Jhevon
Quote:

Originally Posted by hymnseeker
Thanks for the reply, so just let y=0?

1/2x+5/2=0

Does this satisfy what the question wants in asking for ax+by+c=0?

did you read what i said? i said bring everything over to one side, how does that translate to equating y to zero?

$\displaystyle y = \frac 12x + \frac 52$ ............subtract y from both sides

$\displaystyle \Rightarrow 0 = \frac 12x - y + \frac 52$ ...............multiply through by 2

$\displaystyle \Rightarrow 0 = x - 2y + 5$

which is the same thing as

$\displaystyle x - 2y + 5 = 0$

here a = 1, b = -2 and c = 5
• Nov 13th 2007, 10:35 AM
hymnseeker
Thank you for your help. Needed to see it visually - and in answer to your question (although I suspect it was rhetorical), yes, I did read what you said first off.
• Nov 13th 2007, 11:03 AM
Jhevon
Quote:

Originally Posted by hymnseeker
Thank you for your help. Needed to see it visually - and in answer to your question (although I suspect it was rhetorical), yes, I did read what you said first off.

yes, it was rhetorical. you're welcome, and good luck with your class
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