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**sit06603****Member**- Registered: 2005-08-28
- Posts: 2

[Question]

if I have a list of coordinate... let say...

(2, 17)

(3, 25)

(4, 19)

(5, -1)

can u find an equation which can suit all the coordinate above...???

(I suppose... it also will work for more than this number coordinate,

it can be five or more)....

Hopefully there is some one that can help me...!!!

May be I change another way

let say:

x-2y=17 (2, 17)

x-3y=25 (3, 25)

x-4y=19 (4, 19)

x-5y=-1 (5, -1)

Is what my Idea loh... previously I just give a list of

coordinate... and require to produce or figure out an

equation which able to solve the problem... where...

when I....

-> insert 2 in to that equation.... it able to output 17 as

a result....

-> insert 3 in to that equation.... it able to output 25 as

a result....

-> insert 4 in to that equation.... it able to output 19 as

a result....

-> insert 5 in to that equation.... it able to output -1 as

a result....

is that POSSIBLE...??? [ Hopefully is possible].....

plz..... is urgent.... S.O.S~......

An Idea.... just an Idea.....

Now... I assume there is a equation for each coordinate....

where it is in the form of....

-> x-Ay=B (A, B)

-> where, A={0, 1, 2, 3..... n}, and B is the result.....

:. If there is a possible to solve x, y from that list of

equation.... there might be possible.... or any one got

better idea...

plz.... help me..... S.O.S.... is urgent to solve some

problem...... 0_o???

"Plz reply me as soon as possible....." S.O.S~......

Trully,

Wesley

[sit06603@yahoo.com]

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,656

Quick answer: Good News: yes, there are many ways to do it; Bad News: it is a broad and complex field and you need to find a compromise between easy and useful.

The most basic solution is to use a "polynomial":

* with two points you could have an equation like y=ax + b

* with three points you could have an equation like y=ax² + bx + c

* with four points you could have an equation like y=ax³ + bx² + cx + d

The more points you have, the bigger the polynomial, but warning: in between the points and before and after, the line can go wildly up and down

Another solution is to use "best fit" methods. There is something called "least squares regression" that tries to plot a line that gets as close as possible to the points.

There are other more sophisticated methods like "b-splines" and so on.

Try putting the lines into Excel - it has some basic stuff in there.

Tell us some more about what you want to **achieve** and maybe one of us can give you some more specifics.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

I can tell you my method if you want.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

If you graph the four points given, one could assume it might be a parabola opening downward.

If so, then y = ax² + bx + c.

I worked it out and in fact it can be a parabola. So start substituting the points into x and y in y = ax² + bx + c.

Then you have four linear equations with variables a, b, and c.

Play around with those and you get the answer. y=-7x² + 43x - 41 Have fun working it out. If you need more help, just ask!

**igloo** **myrtilles** **fourmis**

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