Last week, in Fundamental Understanding of Mathematics LIX, we took a look at a different kind of multiplication. In an earlier diary, we took an ordered pair, actually three ordered pairs that made up the end points of a triangle, and multiplied them by a number. As expected, the triangle grew larger. Last week we multiplied only one element of the ordered pair, and left the other one alone. And, we multiplied by negative 1. The triangle flipped over.
This week, though, we are going to go back to a question that comes up when we draw triangles using the location of the end points. And that question is: how long are the sides of the triangle?
When we do geometry with a compass and a ruler, we make triangles by drawing lines of a certain length, either by using the ruler, or by setting the compass to draw arcs a certain distance from a point. In either case, we know the length. But what about our Cartesian triangle? We know where the end points are, so we can draw the connecting lines, but how long are those connecting lines?
Let's find out.
Here is a slanted line segment, running from (2, 2) to (10, 6). We can pretend that it is one side of a triangle. When we come up with a method for figuring out how long this side is, we can use it on the other two sides, where ever they are.
Now, since this line is slanted, we can't get the length from the axes of the graph. If the line were horizontal or vertical, finding the length would be a piece of cake, simply subtract one end point from the other. If our line looked like this:
Figuring out the length would be trivial, simply subtract the x coordinates: 10 - 2. This line is 8 units long.
Likewise, if the line were vertical, we would subtract the y coordinates, and find a length of 6 - 2 = 4 units.
Lets put all that on one graph:
The red and blue lines should look familiar. These are the rise (the blue line) and the run (the red line) of our sloped line of unknown length. Taken together, they form a triangle of their own.
Many early mathematicians (land surveyors, probably, we're talking ancient Egyptian early) noticed this relationship and thought to try to develop a way to figure out the unknown length from the two easily known lengths. They succeeded.
These early mathematicians discovered that if you take two squares, one of them with sides the same length as the run, the other with sides the same length as the rise, and added the areas, the sum would be the same as a square with sides of the unknown length.
In this case, the run is 8, so the red square is 64 unit squares. The rise is 4, so the blue square is 16 unit squares. The large gray square, then, is 64 + 16 = 80 square units. My handy calculator says that a square with area 80 has sides 8.94 units long. Of course, the pre-computer age ancients who first discovered this relationship didn't have handy calculators, so they had another problem on their hands: figuring out the sides of a square when you only knew the area. That's a story for another day.
Today, we will merely show that this relationship is true, using a bit of geometry. Let's start with the triangle we made using the rise, run, and unknown length.
Let's extend the rise and run to make squares.
Now, notice that we could make a very large square, with sides run+rise long, by filling in the gaps with triangles the same size and shape as the green triangle.
Now, the area of this larger square is the area of the red square plus the area of the blue square, plus the area of the four green triangles. If I rearrange the triangles, but keep them all within the larger square, the total area should remain the same. Also, the area not covered by triangles should be equal to the area of the red square plus the blue square.
Here is the new arrangement:
Now we only have one square on the inside of the rise + run square, and the sides of the new square are the unknown length, the original sloped line from our graph. (We also have a slight optical illusion, caused by the diagonal lines -- the outside square didn't suddenly tilt. Check it out by scrolling until you can only see the very top or very bottom of the square)
If we take away the triangles from both diagrams, we are left with the red and blue squares on the one hand, and the single gray square on the other. Since the area of the square everything fits inside is the same in both diagrams, and we took away four green triangles from that area (subtracted the same thing) then the area of what's left must be the same as well.
So, the area of the red square plus the area of the blue square must be the same as the area of the gray square.
We can take two lengths, two sides of a right triangle, which we do know, and use this method to figure out the length of the other side of the triangle. Since the other side is the side with a slope (not the horizontal or vertical side), we have a method for finding the length of any sloped line on a graph.
Some of the floor tiles in Mathematics classrooms at UCLA are arranged in this pattern. It's sort of an inside joke. Don't tell them I told you.
Have fun in the comments.