# Thread: Point of best fit?

1. ## Point of best fit?

So we all know that you can apply the "Line of best fit" to a set of points such that the square differences between the estimated values and the real values is minimised but how do you go about finding the "point of best fit" and what does that point describe?

For example I have 3 lines:

$\displaystyle 3x + 4y = 12$
$\displaystyle 3x + 6y = 9$
$\displaystyle x + y = 2$

I want to find the point that intersects all 3 lines. This point clearly does not exist. So instead I want to find the point that is 'closest' to being a triple intersection. I imagine this could be called a 'point of best fit'. How would I go about finding this point?

2. ## Re: Point of best fit?

That's a very interesting problem. I have a couple of questions for you:

1. Does the "point of best fit" have to be on at least one of the lines? On at least two of the lines? Or could it be anywhere?
2. How are you defining "closest"? Since you only have two variables, x and y, in your equations, are you in just the xy plane? If so, are you measuring distance using Euclidean distance?

3. ## Re: Point of best fit?

I see a triangle fromed by the 3 lines and the closet point to the intersection being the middle of the triangle.

4. ## Re: Point of best fit?

Originally Posted by Ackbeet
That's a very interesting problem. I have a couple of questions for you:

1. Does the "point of best fit" have to be on at least one of the lines? On at least two of the lines? Or could it be anywhere?
2. How are you defining "closest"? Since you only have two variables, x and y, in your equations, are you in just the xy plane? If so, are you measuring distance using Euclidean distance?
1. The point can be anywhere at all.
2. Yes, this is only in 2-dimensions with variables x and y. Defining "closest" is the main problem I have with this. What would the point have to satisfy for it to be considered the "closest" to a triple intersection?

I have found a few set of points that satisfy different things:

1. $\displaystyle (1, 17/12)$ is the point such that it has the minimum squared distances from the point to the y coordinate of the lines (ie, its the point on the line of best fit that has the smallest error).
2. $\displaystyle (1, 1)$ is the point such that the summed distance (perpendicular distance) from each line to the point is minimised.
3. $\displaystyle (53/38, 43/38)$ is the point such that the summed squared distance (perpendicular distance) from each line to the point is minimised.
4. $\displaystyle (1, 11/6)$ is the centre of the triangle formed by the 3 lines.

5. ## Re: Point of best fit?

Originally Posted by pickslides
I see a triangle fromed by the 3 lines and the closet point to the intersection being the middle of the triangle.
Which of the myriad definitions of "middle" or "center" of a triangle do you mean? Incenter? Circumcenter? Orthocenter? Centroid? See below: I'm not sure you can pick one center over another until we know the context of the problem.

Originally Posted by Corpsecreate
1. The point can be anywhere at all.
2. Yes, this is only in 2-dimensions with variables x and y. Defining "closest" is the main problem I have with this. What would the point have to satisfy for it to be considered the "closest" to a triple intersection?

I have found a few set of points that satisfy different things:

1. $\displaystyle (1, 17/12)$ is the point such that it has the minimum squared distances from the point to the y coordinate of the lines (ie, its the point on the line of best fit that has the smallest error).
2. $\displaystyle (1, 1)$ is the point such that the summed distance (perpendicular distance) from each line to the point is minimised.
3. $\displaystyle (53/38, 43/38)$ is the point such that the summed squared distance (perpendicular distance) from each line to the point is minimised.
4. $\displaystyle (1, 11/6)$ is the centre of the triangle formed by the 3 lines.
It looks like you've found a number of solutions to your problem. I haven't checked your numbers, but I think that before you choose one solution over any of the others, you need to examine the context of your problem (which wouldn't be a bad idea to post here, actually). Is this problem the same problem, exactly, that you were given? Or have you performed a number of operations on the given problem to reduce it down to what you posted?

6. ## Re: Point of best fit?

I created the problem from a real world situation. The answer to the problem is purely out of interest (it was something that could have been useful to me earlier but not anymore) so it is of no 'real importance'. The problem is as follows:

Suppose you have 2 paints, paint x and paint y. All paints are composed of white paint with added tints of 3 different colours. Call these 3 colours A, B and C. Adding tints does not affect the volume of paint.

if paint x has the composition - A:3 B:3 C:1
and paint y has the composition - A:4 B:6 C:1

Then how much of paint x and y should you add together to yield as close as possible to a composition of A:12 B:9 C:2?

This situation required you to solve the above linear equations however as we know, there is no solution so we are left with finding the 'closest' solution. I am quite capable of finding the 'best point' given that I know what I'm actually looking for. Now that you know the context of the problem, hopefully I could hear your opinion on what the properties of the 'best point' may be.

I'm leaning towards the shortest combined distances from the point to the lines simply because when there IS a solution to the linear equations, this distance is 0. In fact, thinking about it now, the point (1, 17/12) is certainly not what we're looking for since it must be on the line of best fit and that line wont necessarily pass through the triple intersection if there is one. All the other points are still possibilities though.