Lesson 11-2: The Distance Formula (Day 112)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

A = (-3, -2)
B = (6, -14)
AB = ?

To find the length of segment AB, we can use a certain formula:

AB = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
AB = sqrt((6 + 3)^2 + (-14 + 2)^2)
AB = sqrt(9^2 + 12^2)
AB = sqrt(81 + 144)
AB = sqrt(225)
AB = 15

Therefore the desired length is 15 -- and of course, today's date is the fifteenth.

To solve this problem, we used a certain formula known as the Distance Formula. We haven't covered that lesson in the U of Chicago text. Let me check my calendar to see when I'm scheduled to do it...

...and hey, it's today! Today is the day that I'm scheduled to do the Distance Formula. I believe that this is the first time since I started blogging Pappas that a problem related to a certain lesson in the text lands on the exact same day as that actual lesson. (There were some very close calls where the Pappas problem and the relevant lesson were a day or two apart.)

This is what I wrote last year about today's lesson:

Lesson 11-2 of the U of Chicago text is called "The Distance Formula." In the modern Third Edition of the text, the Distance Formula appears in Lesson 11-5.

Let's get to today's lesson. Many students have trouble with graphing throughout Chapter 11, and furthermore, today we learn the Distance Formula, which of course will be difficult for some students.

In the past, I combined Lesson 11-2 with Lesson 8-7, on the Pythagorean Theorem (and indeed, this lesson in the Third Edition is titled "The Pythagorean Distance Formula").

David Joyce has more to say about the Distance Formula:

Also in chapter 1 there is an introduction to plane coordinate geometry. Unfortunately, there is no connection made with plane synthetic geometry. Here in chapter 1, a distance formula is asserted with neither logical nor intuitive justification. Of course, the justification is the Pythagorean theorem, and that's not discussed until chapter 5. In that chapter there is an exercise to prove the distance formula from the Pythagorean theorem. The Pythagorean theorem itself gets proved in yet a later chapter.

Fortunately, the U of Chicago text avoids this problem. Our text makes it clear that the Distance Formula is derived from the Pythagorean Theorem.

Today I post an old worksheet from a few years ago. It introduces the Distance Formula -- but of course, it teaches (or reviews) the Pythagorean Theorem as well  -- including its similarity proof, which is mentioned in the Common Core Standards.

Today is also an activity day, so let's add an activity to this old worksheet. The one Exploration Question given in this lesson is:

22. The distance from point X to (2, 8) is 17.
a. Show that X could be (10, 23).
b. Name five other possible locations of point X. (Hint: Draw a picture.)

Of course, there are infinitely many possible locations for X. The assumption is that students will look for lattice points (points whose coordinates are both integers). These are based on the 8-15-17 Pythagorean triple. Even though we haven't covered vectors yet, it might be helpful for teachers to think in terms of vectors when correcting the student work. Basically, the solutions are to add all versions of the vector <8, 15> to the original point (2, 8). This includes changing one of both signs of <8, 15> as well as switching the x- and y-components. There are seven possible lattice points in addition to the given (10, 23), and students only need to find five of them.

I decided to add today's Pappas problem to the worksheet. It might be years until the next time a Pappas problem lands on the same day as the relevant lesson. So I might as well take full advantage of today's coincidence!

President's Day is on Monday, and so my next post will be Tuesday.