I need to find the midpoint of ST

[mathdad]

The coordinates of the points S and T are S(4, 6) and T(10, 2). If M is the midpoint of line segment ST, find the midpoint of line segment SM.

I need to find the midpoint of ST.
I then need to find the midpoint of SM.

Correct?

 

[Joppy]

Sounds good to me!

 

[mathdad]

Sounds good to me!

I just love to be right.

 

[MarkFL]

Suppose we are given two points in the plane:

\(\displaystyle P_1\left(x_1,y_1\right)\)

\(\displaystyle P_2\left(x_2,y_2\right)\)

The mid-point of the two given points is given by:

\(\displaystyle P_3\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\)

Now, we will find the mid-point of $P_1$ and $P_3$ to be:

\(\displaystyle P_4\left(\frac{x_1+\dfrac{x_1+x_2}{2}}{2},\frac{y_1+\dfrac{y_1+y_2}{2}}{2}\right)=\left(\frac{3x_1+x_2}{4},\frac{3y_1+y_2}{4}\right)\)

Suppose we consider $P_3$ to be the first mid-point, and $P_4$ to be the second mid-point...which we will label $M_1$ and $M_2$ respectively. What do you suppose is the formula for the $n$th mid-point $M_n$? What would you say $M_n$ approaches as $n$ goes to infinity?

 

[mathdad]

Suppose we are given two points in the plane:

\(\displaystyle P_1\left(x_1,y_1\right)\)

\(\displaystyle P_2\left(x_2,y_2\right)\)

The mid-point of the two given points is given by:

\(\displaystyle P_3\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\)

Now, we will find the mid-point of $P_1$ and $P_3$ to be:

\(\displaystyle P_4\left(\frac{x_1+\dfrac{x_1+x_2}{2}}{2},\frac{y_1+\dfrac{y_1+y_2}{2}}{2}\right)=\left(\frac{3x_1+x_2}{4},\frac{3y_1+y_2}{4}\right)\)

Suppose we consider $P_3$ to be the first mid-point, and $P_4$ to be the second mid-point...which we will label $M_1$ and $M_2$ respectively. What do you suppose is the formula for the $n$th mid-point $M_n$? What would you say $M_n$ approaches as $n$ goes to infinity?

I like the fact that you always reply to my questions but please keep it simple with me. I took precalculus in 1993. It is a long time ago. I have no clue what M_n tends to as n goes to infinity. I am not too familiar with the concept of infinity. I hope to be there someday but for now, please elaborate.

 

[MarkFL]

I like the fact that you always reply to my questions but please keep it simple with me. I took precalculus in 1993. It is a long time ago. I have no clue what M_n tends to as n goes to infinity. I am not too familiar with the concept of infinity. I hope to be there someday but for now, please elaborate.

I'm really just asking you a conceptual question here...think about what happens as you find the mid-point of a line segment, then you find the mid-point of the half having the first point as an end-point, and you keep repeating this over and over...where do you think the $n$th mid-point is heading?

Do you have any hypothesis for the formula of the $n$th mid-point $M_n$?

 

[mathdad]

I'm really just asking you a conceptual question here...think about what happens as you find the mid-point of a line segment, then you find the mid-point of the half having the first point as an end-point, and you keep repeating this over and over...where do you think the $n$th mid-point is heading?

Do you have any hypothesis for the formula of the $n$th mid-point $M_n$?

The nth midpoint is heading to 1 or 0.

 

[MarkFL]

The nth midpoint is heading to 1 or 0.

Imagine you have a piece of string anchored at one end to a sheet of wood. Now you cut away half the string and discard the free end. You do this over and over...and you notice that the free end is getting closer and closer to the anchored end. If we do this an infinite number of times, we can image the free end becoming indistinguishable from the anchored end. So, we should expect (using the labels from my first post):

\(\displaystyle \lim_{n\to\infty}M_n=P_1\)

Now, if you go back to my first post, you should observe that we could hypothesize:

\(\displaystyle M_n=\left(\frac{\left(2^n-1\right)x_1+x_2}{2^n},\frac{\left(2^n-1\right)y_1+y_2}{2^n}\right)\)

Let $n=1$ and $n=2$ and you'll see that it agrees with the mod-point formulas derived in my post.

I know you haven't gotten to the chapter on mathematical induction yet, but using that technique, we can prove the above is true.

Notice that this is algebraically equivalent to:

\(\displaystyle M_n=\left(\left(1-2^{-n}\right)x_1+2^{-n}x_2,\left(1-2^{-n}\right)y_1+2^{-n}y_2\right)\)

Now, using the fact that:

\(\displaystyle \lim_{n\to\infty}2^{-n}=0\)

We see that:

\(\displaystyle \lim_{n\to\infty}M_n=\left(x_1,y_1\right)=P_1\)

This is the math agreeing with intuition. :D

 

[mathdad]

Imagine you have a piece of string anchored at one end to a sheet of wood. Now you cut away half the string and discard the free end. You do this over and over...and you notice that the free end is getting closer and closer to the anchored end. If we do this an infinite number of times, we can image the free end becoming indistinguishable from the anchored end. So, we should expect (using the labels from my first post):

\(\displaystyle \lim_{n\to\infty}M_n=P_1\)

Now, if you go back to my first post, you should observe that we could hypothesize:

\(\displaystyle M_n=\left(\frac{\left(2^n-1\right)x_1+x_2}{2^n},\frac{\left(2^n-1\right)y_1+y_2}{2^n}\right)\)

Let $n=1$ and $n=2$ and you'll see that it agrees with the mod-point formulas derived in my post.

I know you haven't gotten to the chapter on mathematical induction yet, but using that technique, we can prove the above is true.

Notice that this is algebraically equivalent to:

\(\displaystyle M_n=\left(\left(1-2^{-n}\right)x_1+2^{-n}x_2,\left(1-2^{-n}\right)y_1+2^{-n}y_2\right)\)

Now, using the fact that:

\(\displaystyle \lim_{n\to\infty}2^{-n}=0\)

We see that:

\(\displaystyle \lim_{n\to\infty}M_n=\left(x_1,y_1\right)=P_1\)

This is the math agreeing with intuition. :D

Very impressive.

 

[HallsofIvy]

About as simply as I can put it: one half of one half is one fourth! If M is halfway from S to T then half way from S to M is one fourth of the way from S to T.

 

[mathdad]

Thank you everyone.