The Line

Eric Kosslyn showed me this:...

Generating Pythagorean triplets:
Let N be the set of natural numbers.
Let a^1=a and let a^(n+1) = (a^n)(a). Exponentiation.
Identity: (x^2-y^2)^2 +(2xy)^2=(x^2+y^2)^2 (work it out if you do not see it).
For all x and y in N, x+y is in N and xy is in N, so
For all x and y in N
If a=x^2-y^2 and b=2xy and c=x^2+y^2 you have a,b,c, in N and...
a^2+b^2=c^2. Pythagorus.

We can use this to generate all pythagorean triplets, and then we can use pythagorean triplets to generate systems of simultaneous equations in two variables which come out "neat".

Example:
<3,4,5> and <5,12,13> are pythagorean triplets.
Let m(L1), the slope of line 1, be 4/3 and let m(L2), the slope of line 2, be -5/12.
Use <3,4> (from <3,4,5>).
Use <-12,5> (from 5,12,13>)
Take common multiples of the first and second elements: 48 is a multiple of both 3 and -12, 40 is a multiple of both 4 and 5.
Use p<48, 40> as the point of intersection. We let L1 and L2 intersect at the point p<48, 40>.

Then, from the vector form of the equation for the line we have...
for all p(x,y) in L1, there exists an r in R such that
(x,y) = <48,40> + r<3,4>
and
for all p(x,y) in L2, there exists r in R such that
(x,y)= <48,40> + r<-12,5>

Read "a => b" as "a implies b", or "a, so b".

Find two integer-valued points in L1:
r=6 => (x,y) = <48,40> + 6<3,4> => (x,y) = <48,40> + <18,24> => (x,y) = <66, 64>.
r=-8 => (x,y) = <48,40> + -8<3,4> => (x,y) = <48,40> + <-24,-32> => (x,y) = <24, 8>.
So {p<66,64>, q<24,8>} is a subset ot L1.

Find two integer points in L2:...
r=-9 => (x,y) = <48,40> + -9<-12,5> => (x,y) = <48,40> + <108,-45> => (x,y) = <156,-5>.
r=11 => (x,y) = <48,40> + 11<-12,5> => (x,y) = <48,40> + <-132,55> => (x,y) = <-84,95>.
So {s<156,-5>, t<-84, 95>} is a subset of L2.

I usually use p1, p2, p3, p4 with subscript 1,2,3,4, but I don't know how to do this in the Blogger fonts . It would be {p1(x1,y1), p2(x2,y2)} subset L1 ("subset" is a U on its side).

An n-dimensional cartesean coordinate system is a function from Euclidean n-space to R^n which is one-to-one, onto, and which preserves the relation "between". Make an analogy with a proper assignment of names to people in a group: each person gets a name, each person gets only one name, no name applies to more than one person.

Let m(pq) be the slope of the segment (pq) (this is not the proper notation, but I don't know how to make the superscript bar over "pq" in Blogger fonts). Slope is a function from intervals to real numbers.
Let d(p,q) be the distance from p to q. Distance is a function from pairs of points to real numbers.
Let u(pq) be the midpoint of the segment pq. Midpoint is a function frrom intervals to points.

Repetition of "function" in this presentation prepares students for later work, where functions become a topic of study in abstract. Students will find this much easier if they have encountered functions before. Writing a Math curriculum shares some features with writing a murder mystery. You cannot have the Great Detective tell the assembled suspects in the concluding chapter: "The butler did it" if the butler never appeared earlier. He has to walk on and serve cucumber sandwiches in chapter 3.

Exercise:...
...........................................................................................................................................................
Let T subset of E^2XR^2 be a cartesian coordinate system.
Let L1 and L2 be lines in E^2 such that

{p<66,64>, q<24,8>} is a subset ot L1
and
{s<156,-5>, t<-84, 95>} is a subset of L2.

Find:...

m(pq)=___
d(p,q)=___.
u(pq)=___
m(st)=___
d(s,t)=___
u(st)=___
point-slpoe form of L1:_______________________
slope-intercept form of L1: ____________________
intercept form of L1: _________________________
Standard form of L1: ________________________
vector form of L1: ___________________________
matrix (determinant) form of L1:__________________

point-slpoe form of L2:_______________________
slope-intercept form of L2: ____________________
intercept form of L2: _________________________
Standard form of L2: ________________________
vector form of L2: ___________________________
matrix (determinant) form of L2:__________________

L1 intersect L2 =___________________ ("intersect"" is a U upside down).
By substitution.
By matrix elimination.
By Cramer's rule.
............................................................................................................................................

That last answer should look like this..."L1 intersect L2 = {p<48,40>}".

Because we used pythagorean triplets for the slope and integers for the multipliers in generating points, the distance comes out integer. Because we used BOTH odd or BOTH even integers for the multipliers in generating points, the midpoint comes out integer. Because we built this exercise from the point of intersection p<48,40> to start, this comes out integer.

Once people know how to add and subtract rational numbers, (by third grade if their parents are doing their job), they are ready to start basic Analytic Geometry of intervals in one-space and lines in 2-space (which schools commonly call "Pre-Algebra" and "Algebra I").

I teach in the following sequence:
1) Linear equations in one variable
2) Linear inequalities in one variable (Let T subset of E^1XR^1 be a one-dimensional cartesian coordinate system).
3) Intervals in 1-space (inequalities with absolute value).
4) Slope of intervals in 2-spae.
5) Linear equations in 2-space.

US textbooks usually save graphing linear inequalities of one variable involving absolute value to the course we call Algebra II. This misses the opportunity to derive the equation of the line in 2-space from the definitions of "interval" and "slope".

Since the line between the points a and b is the set of all points p such that m(ap) = m(ab), the point-slope form of the equation of the line falls out of the defintition of the line.