Question Corner and Discussion Area

I have several questions. First, why does y=mx+b work?Secondly, how do you prove that something is an overdetermined system through graphing?

Third, How do you prove that three lines which you can SEE are not collinear actually aren't?

- I'm not entirely sure what you mean by "work" in your first
question; let me assume that what you mean is "why can all lines
(except for vertical lines) always be described by an equation of
the form y = mx + b?"
One elementary way to see it is to look at the picture below. If L is a non-vertical straight line, then the triangles below are similar triangles. The ratio V/U is the same as the ratio v/u. We call this ratio the

*slope*m of the line.* L* | * | * | V * v | * | | *<- u ->+---- <---- U --->

(For example, if U is three times u, then the picture below shows why V has to be three times v, so V/U = (3v)/(3u) = v/u. Similar reasoning can be used for the general case).

* * | | * v | * | | *-- u -- | * | | * v | L* | V=3v *-- u -- | * | | * v | * | | *<- u -> <-------- U=3u -------->

If we draw coordinate axes and let b be the value of y where the line passes through the y-axis, then we can ask ourselves, "what are the mathematical conditions that determine whether or not a point (x,y) lies on the line?"

Looking at the picture below, we see that y-b is the "V" of the above picture and x is the "u", so the ratio (y-b)/x equals m, in other words, y-b = mx, so y = mx + b.

y| * | L* | | * |<- y-b | * | | * | b*--------- * | | ---+----------------- 0 x

That is why all lines (except for vertical lines, whose equation is of the form x = constant) can be expressed in the form y=mx + b.

- In general, graphing isn't a means of
*proving*something is an overdetermined system, but rather a way get to get some visual insight into what the concept means.It really all depends on what you mean by "something". Are you talking in general, or are you referring to something specific like a set of linear equations?

Without knowing the details of the context in which you're asking the question, let me just make some general comments.

If you have a collection of equations, each of which has its solution set, then the set of points that satisfy

*all*the equations is the intersection of these solution sets.For example, if you have the two equations x + y = 1 and 2x + y = 3, the solution set to the first equation is the line y=1-x, and the solution set to the second equation is the line y=3-2x. The set of points (x,y) that satisfy both the equations is the intersection of those two lines (which is the single point (2,-1)).

Now, if you start with just one equation, chances are the solution set has a whole range of points in it rather than just being a finite number of isolated points. For instance, the solution set to the equation x+y=1 is an entire line.

However, as you add more equations, you are intersecting the solution set with other solution sets, so you get something smaller. A key question is, how many equations does it take until you get down to just a finite number of isolated points?

If your equations contain just two variables, then it generally takes two equations. The solution set to each equation is usually something like a line or a curve, and when you intersect them, you get a finite a number of points.

If your equations contain three variables, it generally takes three equations. The solution set to each equation will be something like a plane or other surface. Intersecting two of those gives you a curve, but intersecting that curve with a third surface will get you down to individual points. (For example, two adjacent faces of a cube will intersect along an entire edge, but three adjacent faces of a cube will intersect in a single point, namely a corner vertex).

And so on: if you have n equations, it generally takes n equations to reduce the solution set down to a single point or a collection of finitely many points.

(I say "generally" because you really need n

*different*equations, and you may not always have this. For example, the two equations y=x+1 and 2x = 2y - 2 both have the same solution set, so specifying both of these equations together is no different from specifying just one of them. In this case, you'd need to add a third equation to get down to just a single point for the solution).A collection of equations is

*overdetermined*if it contains*more equations*than are needed to get the solution set down to a single point or finite collection of points.For example, if you take the equations for three lines in a plane, they form an overdetermined system because just two lines are enough to get you a single intersection point. The third line might not even pass through that intersection point at all (in which case, there is no solution to the system of the three equations), or it might pass through it (in which case the third equation is redundant and could have been dropped altogether).

Note that an overdetermined system of equations usually has no solution. It is only in special cases that a solution exists. For example, the equations for three lines in a plane will have a common solution only if the lines are concurrent; generally, there will be no point that lies on all three lines.

By contrast, an underdetermined system usually does have a solution. It is only in special cases that no solution exists. For example, the equations for two lines in a plane will have a common solution in all cases except when the lines are parallel, for as long as the lines aren't parallel there will be an intersection point.

- I assume you mean "cuncurrent" in your question (collinearity
is a property of
*points*, meaning they all lie on one line, concurrency is a property of lines, meaning they all pass through a single point).To prove three lines are not concurrent, you prove that there is no pair of numbers (x,y) which satisfies all three equations. One way to do that is to solve the first pair of equations and discover that there's only one solution (the point of intersection of the first two lines), then prove that this solution does not satisfy the third equation (i.e., that the intersection point of the first two lines does not lie on the third line).

For example, suppose you have the lines y=x+1, y=2x+3, and y=3x+2. The intersection of the first two lines is given by y = x+1 = 2x+3, so x+1 = 2x+3, so 0 = x+2 (subtracting x+1 from both sides), so x=-2, and y=(-2)+1 = -1.

Therefore, (-2,-1) is the point of intersection of the first two lines. But (-2,-1) does not satisfy the equation of the third line, since -1 does not equal 3(-2) + 2. This proves that the lines are not concurrent.

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