Hello,

I have come across this problem and need help solving it. Give the equation of the line from (3, -3) to (-1, 3) in the y=mx+c format. (Headbang)

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- Apr 21st 2012, 09:49 PMjabelone[SOLVED] Help with y=mx+c
Hello,

I have come across this problem and need help solving it. Give the equation of the line from (3, -3) to (-1, 3) in the y=mx+c format. (Headbang) - Apr 21st 2012, 10:20 PMbiffboyRe: Help with y=mx+c
m is the gradient of the line. So m= (3 minus -3) divided by (-1 minus 3) When you have found m put it into y=mx+c. Now substitute into this x=3 and y=-3 (because you know this point fits the line) and you will get c.

- Apr 21st 2012, 10:30 PMjabeloneRe: Help with y=mx+c
thanks,

So would this be right: y =^{-}1.5 x 3 + 3

Thank-you for your help, I couldn't find it in my notebook. - Apr 21st 2012, 11:47 PMbiffboyRe: Help with y=mx+c
You are correct that m= -1.5 So equation is y= -1.5x + c Now (3,-3) is on this line so -3= (-1.5*3)+c which gives c=1.5

So equation of line is y= -1.5x+1.5 - Apr 22nd 2012, 12:18 AMjabeloneRe: Help with y=mx+c
Thank you very much.

- Apr 22nd 2012, 05:22 AMWilmerRe: Help with y=mx+c
Another way.

Once you get the slope of -3/2, you need to find c in (0,c).

Use one of given poinrs; say (3,-3):

(-3 - c) / (3 - 0) = -3/2

-6 - 2c = -9

c = 3/2