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HELP, please i have a test tomorrow - -
tofumonzter
post Nov 16 2004, 12:48 AM
Post #1


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so mr/mrs math please help me on this. i have a test tomorrow. and if you can help me figure this out i'll be your slave for one day or anything. pleaseeee~ pinch.gif

ok this is what i need to know.

i need to know how to

convert y=mx+b into (y2-y1)=m(x2-x1) and then (ax+by)=c
basicly it's slope-intercept form into point-slope form and then standard form

PROBLEM : slope 2; y-intercept 5

please all the smart people that knows good math help me on this, and i'll be your slave for one day pinch.gif pinch.gif

PLEASE pinch.gif
 
Spirited Away
post Nov 16 2004, 01:17 AM
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Quand j'étais jeune...
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I think this website or this cheat sheet might help.


Just look around online. There are a lot of examples that may help you study.
 
whomps
post Nov 16 2004, 01:20 AM
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I just reread this right now and find out that it's REALLY confusing. But it took me awhile to type it so blah. Meh. What are you taking right now?

Y = mx + b
Y is the Y, m is the slope, x is x, and be is the y intercept.

If they give you the equation Y = mx + b, they give you the slope and the y-intercept. You can already find another point on the line by using the the y-intercept and slope.

The y-intercept already gives you a point..>> (0, y)

(y2 - y1) = M
(x2 - x1)

^ To find another point.

So when you have y2-y1 = m(x2-x1), you substitute in your two points.
A point would be (x1, y1) and (x2, y2)

Okay so you found the two points, LETS PRETEND they're uhm (4, 1) and (6,3).
to get them into (ax+by=c) form, you have to substitute them.
Soo substituing them, you would receive two equations because there are two points.

First equation >> 4a+ 1b = c ---- Now you have a system of equations! =o
Second equation >6a + 3b = c

To solve the system of equations, multiply the first equation by (-3)

-12a + -3b = -3c
6a + 3b = c

Now add them. Straight down. To get one equation..

-6a = -2c << Divide (-2) by both sides
To get..
3a = c

Now, go back to the FIRST equation.

4a + 1b = c

Substitute C

4a + 1b = 3a

Solve the equation. Substract 3a from both sides,

a + b = 0

Change ^ that to (b = -a)

Uhm.. and your original equation

ax + by = c

Substitute. B and C

ax + (-a)y = 3a

Divide all sides by A.

x + (-y) = 3

There's an easier way to do it I think.. because m = -a/b but uhm yeah.
Got the last part (ax + by = c) from this one page..
 
tofumonzter
post Nov 16 2004, 01:31 AM
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thanks eve and loan, you two are such a awesome poop to me. tongue.gif biggrin.gif
 

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