Repost: Adding and Subtracting Integers
This is, without a doubt, THE most popular post at Text Savvy. I re-post it now with this request: Can we please introduce vectors in late-elementary/early middle-school math? Please?
One of the reasons we get these migraines over integers is that, at least up to the point that we as students were actually introduced to operations with negative numbers, we had been taught (correctly) that addition is an operation that describes combination and subtraction describes extraction. We know, for instance, that adding values is like combining collections of objects, and subtracting values is like removing a collection of objects from another collection.
Then we get to integer math, at which point we are asked, judging by present-day treatments in textbooks, to understand the idea that we should be able to, for example, add a "negative collection" to another "negative collection." Or we must throw away and disregard as ridiculous all that "collection" talk.
Mathematics is always described as a beautifully and rigorously universal subject in every detail--when an idea is laid down and proven in mathematics, it applies everywhere and always. But, to my mind, this is not the way mathematics works, and people should really stop spreading this "universal" rumor.
When you add or subtract with integers, you are NOT combining collections or extracting from collections; you are "moving" in certain directions.
As an example, below is a (poorly drawn) submarine at a depth of -5 whatevers. This doesn't mean that depth can be negative (how could it?); it simply means that the submarine's position, in relation to a number line, can be described as -5.
If we add 3, we are adding buoyancy. Because 3 is positive, we are adding positives. Thus, the sub goes up.
If the sub starts at -5, and we subtract 3, however, we are "getting rid of" positives. We are "getting rid" of When we get rid of positives, the sub must sink:
Finally, let's find -5 - (-3). Here we are "getting rid of negatives." When we get rid of negatives, the sub must go up.
Then we get to integer math, at which point we are asked, judging by present-day treatments in textbooks, to understand the idea that we should be able to, for example, add a "negative collection" to another "negative collection." Or we must throw away and disregard as ridiculous all that "collection" talk.
Mathematics is always described as a beautifully and rigorously universal subject in every detail--when an idea is laid down and proven in mathematics, it applies everywhere and always. But, to my mind, this is not the way mathematics works, and people should really stop spreading this "universal" rumor.
When you add or subtract with integers, you are NOT combining collections or extracting from collections; you are "moving" in certain directions.
As an example, below is a (poorly drawn) submarine at a depth of -5 whatevers. This doesn't mean that depth can be negative (how could it?); it simply means that the submarine's position, in relation to a number line, can be described as -5.
If we add 3, we are adding buoyancy. Because 3 is positive, we are adding positives. Thus, the sub goes up.
If the sub starts at -5, and we subtract 3, however, we are "getting rid of" positives. We are "getting rid" of When we get rid of positives, the sub must sink:
Finally, let's find -5 - (-3). Here we are "getting rid of negatives." When we get rid of negatives, the sub must go up.
Labels: education, mathematics
"Interesting" Is not Subjective
How could it be? Anyone with half a brain can see that this is a really poor attempt to interest kids in math.
Far more interesting are videos of water tanks slowly filling up. Or maybe paint drying.
Far more interesting are videos of water tanks slowly filling up. Or maybe paint drying.
18- and 72-Degree Rotations
And now, finally onto 18° and 72° rotations. Once again we can get some inspiration from working with triangles.
The 36-72-72 triangle is an interesting one. Take a look at Triangle ABC at the right. Angle A is a 36° angle, angle B is a 72° angle (36 + 36 = 72), and angle C is a 72° angle, so this triangle is a 36-72-72 triangle. And it's an isosceles triangle, with AB congruent to AC.
When we draw line segment BD from point B to side AC, bisecting angle B, we create two similar triangles: Triangle BCD is also a 36-72-72 triangle, so it is similar to Triangle ABC.
This, of course, means that the two triangles' corresponding sides are proportional.
Since Triangle BCD has two congruent angles, it is an isosceles triangle, and the sides opposite these congruent angles are also congruent. Thus, side BC and side BD are congruent, as shown in the diagram at the left.
Interestingly, side AD is also congruent to sides BC and BD because Triangle ADB is also an isosceles triangle (a 36-36-108 triangle). Since BD is congruent to BC, and BD is also congruent to AD, then AD is congruent to BC.
Now we get to do something that mathematicians do all the time--we get to make stuff up and watch people wonder why.
So, let's just make up some values for two of the segments in this diagram: let's say that the length of DC is 1 and the length of BC is x. This of course means that BD and AD each have a length of x as well.
The values that we assign to these lengths is not terribly important, so long as we don't assign values that can't be true. (So, we can't say that both DC and BC are x, for example, because we know that these segments are not congruent.)
Since we know that Triangle BCD and Triangle ABC are similar triangles, and thus their corresponding sides are proportional, we can write a proportion to make the relationship between the sides a little clearer.
So, AB = x2. Not very interesting. But look at side AC. We can describe its length as x + 1, whereas side AB has a length of x2. Remember that Triangle ABC is an isosceles triangle with side AB congruent to side AC. So, because these two lengths are congruent, this equation must be true:
x2 = x + 1
Now that is interesting. The value of x is such that when you square it, you obtain the same value as when you add 1 to it. What is this value? You can see how it is determined at the right.
(If you don't remember, or you haven't seen it before, the third equation at right comes from a technique for solving quadratic equations called completing the square.)
More importantly, though, this number is called the golden ratio. It is approximately equivalent to 1.618034, and it has its own special symbol, &Phi, which is spelled out as "phi" and pronounced "fee."
Okay, so back to our triangle. We can draw a segment which bisects angle A and bisects side BC at a 90° angle. This gives us an 18-72-90 right triangle, which we can work with to determine the formulas for 18° and 72° rotations. (We just have to imagine that the triangle is rotated left, with the 18° angle sitting at the origin of the coordinate plane.)
Recall that x is equal to this special number we are now writing as &Phi, so our hypotenuse, side AC, has a length of &Phi + 1 (which is the same as &Phi2!), and our vertical side has a length of &Phi/2.
We can make these values a little easier on the eyes by dividing both lengths by &Phi/2. This of course gives us 1 for the vertical side, and for the hypotenuse (remember that &Phi + 1 = &Phi2 and that dividing a fraction is the same as multiplying its reciprocal):
All that's left is the horizontal side. The Pythagorean Theorem makes quick work of it (remember again, &Phi + 1 = &Phi2).
Now we have the triangle and formulas we were after for our 18° and 72° rotations:
As always, there's the helpful more that has come before: here, here, here, here, here, and here.
The 36-72-72 triangle is an interesting one. Take a look at Triangle ABC at the right. Angle A is a 36° angle, angle B is a 72° angle (36 + 36 = 72), and angle C is a 72° angle, so this triangle is a 36-72-72 triangle. And it's an isosceles triangle, with AB congruent to AC.
When we draw line segment BD from point B to side AC, bisecting angle B, we create two similar triangles: Triangle BCD is also a 36-72-72 triangle, so it is similar to Triangle ABC.
This, of course, means that the two triangles' corresponding sides are proportional.
Since Triangle BCD has two congruent angles, it is an isosceles triangle, and the sides opposite these congruent angles are also congruent. Thus, side BC and side BD are congruent, as shown in the diagram at the left.
Interestingly, side AD is also congruent to sides BC and BD because Triangle ADB is also an isosceles triangle (a 36-36-108 triangle). Since BD is congruent to BC, and BD is also congruent to AD, then AD is congruent to BC.
Now we get to do something that mathematicians do all the time--we get to make stuff up and watch people wonder why.
So, let's just make up some values for two of the segments in this diagram: let's say that the length of DC is 1 and the length of BC is x. This of course means that BD and AD each have a length of x as well.
The values that we assign to these lengths is not terribly important, so long as we don't assign values that can't be true. (So, we can't say that both DC and BC are x, for example, because we know that these segments are not congruent.)
Since we know that Triangle BCD and Triangle ABC are similar triangles, and thus their corresponding sides are proportional, we can write a proportion to make the relationship between the sides a little clearer.
So, AB = x2. Not very interesting. But look at side AC. We can describe its length as x + 1, whereas side AB has a length of x2. Remember that Triangle ABC is an isosceles triangle with side AB congruent to side AC. So, because these two lengths are congruent, this equation must be true:
Now that is interesting. The value of x is such that when you square it, you obtain the same value as when you add 1 to it. What is this value? You can see how it is determined at the right.(If you don't remember, or you haven't seen it before, the third equation at right comes from a technique for solving quadratic equations called completing the square.)
More importantly, though, this number is called the golden ratio. It is approximately equivalent to 1.618034, and it has its own special symbol, &Phi, which is spelled out as "phi" and pronounced "fee."
Okay, so back to our triangle. We can draw a segment which bisects angle A and bisects side BC at a 90° angle. This gives us an 18-72-90 right triangle, which we can work with to determine the formulas for 18° and 72° rotations. (We just have to imagine that the triangle is rotated left, with the 18° angle sitting at the origin of the coordinate plane.)Recall that x is equal to this special number we are now writing as &Phi, so our hypotenuse, side AC, has a length of &Phi + 1 (which is the same as &Phi2!), and our vertical side has a length of &Phi/2.
We can make these values a little easier on the eyes by dividing both lengths by &Phi/2. This of course gives us 1 for the vertical side, and for the hypotenuse (remember that &Phi + 1 = &Phi2 and that dividing a fraction is the same as multiplying its reciprocal):
All that's left is the horizontal side. The Pythagorean Theorem makes quick work of it (remember again, &Phi + 1 = &Phi2).Now we have the triangle and formulas we were after for our 18° and 72° rotations:
As always, there's the helpful more that has come before: here, here, here, here, here, and here.
Labels: mathematics
Slopes of Perpendicular Lines
Why do the equations for perpendicular lines have slope values that are negative reciprocals of each other?
Easy. The formula for the 90° rotation of any point (x, y) about the origin can be written as (-y, x):
When we rotate an entire line 90°, we rotate all of the points 90°. The original line has a slope that can be described using any two of its points (x1, y1) and (x2, y2) with this formula:
When we rotate those points 90°, we have to replace every x with a -y and every y with an x, because, as we mentioned, the formula for a 90° rotation is (-y, x), which gives us
or, rewritten:
which is the reciprocal of the original slope formula multiplied by negative 1.
This of course works for 270° rotations (y, -x) as well, because we just end up moving the negative sign from the denominator to the numerator.
Easy. The formula for the 90° rotation of any point (x, y) about the origin can be written as (-y, x):
When we rotate an entire line 90°, we rotate all of the points 90°. The original line has a slope that can be described using any two of its points (x1, y1) and (x2, y2) with this formula:
When we rotate those points 90°, we have to replace every x with a -y and every y with an x, because, as we mentioned, the formula for a 90° rotation is (-y, x), which gives us
or, rewritten:
which is the reciprocal of the original slope formula multiplied by negative 1.
This of course works for 270° rotations (y, -x) as well, because we just end up moving the negative sign from the denominator to the numerator.
Labels: mathematics
Cinemathematics, the Sequel
It becomes increasingly difficult over time to offer unique, novel responses to the same old stories that we hear over and over in education.
Here's Cocktail Party Physics with one such story:
The American students got these questions wrong because they learn from us throughout their schooling that something like 6 + 9 means "9 happens to 6 to make 15." So, their understanding of the addition looks like this:
This representation may look a little odd, but, as far as our teaching is concerned, all the important parts are there. That is, 9 happened to 6, and the result is 15. The expression 6 + 9 is just a means to get the answer we're after. Once we've got that result, the other numbers become part of the past. The addends are just flat rocks spaced out nicely across a river. Students jump to the first one, then to the second one, and then, taking their cue from the equals sign, jump to the opposite bank of the river with their answer.
Imagine how frightening it must be for students who are educated in this manner to see 6 + 9 = __ + 4! Some students will see the equals sign and just jump, because they don't know any better, so they'll write 15 in the blank. Others may be able to tell that the opposite bank is farther away than normal but won't have a clue what to do next, so they'll just guess. Still others may see the "+ 4" sign in the distance and deduce that someone put it there as a hint for what to do next, so they'll write 19 in the blank.
And we haven't even discussed those students who don't know how to jump from rock to rock accurately in the first place. All of the errors described above come from students (who are taught by teachers) who KNOW HOW TO ADD!
I suppose I should go read the paper now.
Update: Pffh. I don't even need to read this stuff anymore to know what's in it. But I will anyway.
Here's Cocktail Party Physics with one such story:
A study by Capraro, et al in Psychological Reports (106(1), 49-53 (2010)), which draws on their previous data in Li, et al. (Cognition and Instruction, 26, 195-217 (2008) compares 6th grade students from different countries. . . .
The results from the first two questions I posed from their study (6 + 9 = __ + 4 and __ + 8 = 12 + 5) were surprising/appalling. Only 28.6% of American students got these questions right. The Chinese and Korean rates were in the 90+% range and the Turkish rates were 61% and 79% respectively. . . .
For the "Type C" problem __ + 3 = 5 + 7 = __, American students got the first blank right 23.8% of the time, while the rates for other students were 98.6% (Chinese), 86.5% (Korean) and 60.2% (Turkish). Interestingly, the correct rates for the second blank were much more comparable: 86.7% (American), 97.9% (Chinese), 93.3% (Korean) and 86.0% (Turkish).
- The sixth-grade American students who got these questions wrong were or are mathematical morons.
- The little gomers did not get that way all by themselves.
- Our failure as educators that these results highlight ultimately has nothing to do with our lack of knowledge about math or our lack of experience teaching. We "know" lots of math, and we impart this knowledge relatively well; we just impart the wrong math.
- Traditionalist ideologues will argue that these results are to be expected, since students nowadays are taught to just guess and check their answers and have no experience solving basic problems like these. These ideologues are wrong.
- Reform ideologues will argue that these results are to be expected, since students are still taught rote procedures without meaning or motivation. These ideologues are also wrong.
Fundamental to mathematics is the idea of a relation, which is a formal way of stating that two objects are related in some specific way. For instance, the object "2 + 3" is related to the object "5" by the equality. This notion, however, can hide that something very important has occurred. We have taken a conceptual process, addition, and restated it in terms of a statement about static relations. No matter what I do, I cannot break the relation "2 + 3 = 5." By contrast, if I constrain myself to thinking about the addition process, then it is harder to separate that statement about Platonic ideals from the perhaps imperfect implementation of the addition process. The equality relation, then, tells us about what is.
The American students got these questions wrong because they learn from us throughout their schooling that something like 6 + 9 means "9 happens to 6 to make 15." So, their understanding of the addition looks like this:
This representation may look a little odd, but, as far as our teaching is concerned, all the important parts are there. That is, 9 happened to 6, and the result is 15. The expression 6 + 9 is just a means to get the answer we're after. Once we've got that result, the other numbers become part of the past. The addends are just flat rocks spaced out nicely across a river. Students jump to the first one, then to the second one, and then, taking their cue from the equals sign, jump to the opposite bank of the river with their answer.
Imagine how frightening it must be for students who are educated in this manner to see 6 + 9 = __ + 4! Some students will see the equals sign and just jump, because they don't know any better, so they'll write 15 in the blank. Others may be able to tell that the opposite bank is farther away than normal but won't have a clue what to do next, so they'll just guess. Still others may see the "+ 4" sign in the distance and deduce that someone put it there as a hint for what to do next, so they'll write 19 in the blank.
And we haven't even discussed those students who don't know how to jump from rock to rock accurately in the first place. All of the errors described above come from students (who are taught by teachers) who KNOW HOW TO ADD!
I suppose I should go read the paper now.
Update: Pffh. I don't even need to read this stuff anymore to know what's in it. But I will anyway.
Falkner, Levi, and Carpenter (1999) reported from their investigation in a single school of the problem "8 + 4 = a + 5", all 145 sixth-graders filled the box with 12 or 17. . . . According to Carpenter, Franke, and Levi (2003), students may have three different misconceptions of the equal sign: equal sign may mean "the answer comes next" ignoring the rest of the problem (p. 10), that is, 8 + 4 = [12]; students may "use all the numbers" (p. 11) such as 8 + 4 + 5 = 17, arbitrarily restructuring the sentence; or they may put 12 in the box and "extend the problem" (p. 11) as 8 + 4 = [12] + 5 = 17.
Labels: education, mathematics, research
75- and 15-Degree Rotations
In this post, we get to what is close to the end of how far our initial perspectives can take us with regard to determining rotation formulas.
To recap, we have derived formulas for 90° (180°, 270°), 45°, and 30° rotations (Click on the degree measures to go to the relevant post. Click on the pics to enlarge):
90°:
(-y, x)
45°:
30°:
The pattern to note above--the one I asked about in the previous post--is that the hypotenuses of the triangles can be found in the denominators in the formulas, and the other side lengths of the triangles can be found in the coefficients of x and y in the formulas.
In fact, for both 30° and 45° angles, the rotation formulas can be written as follows:
where, for the time being, opposite side means "the side that is directly opposite the rotation angle in question" and adjacent side means "the side that is not the hypotenuse or the opposite side."
Though we don't really know it yet, this is the formula for any rotation about the origin. That is, if we replace the coefficients above with their somewhat more abstract and useful equivalents, sine (sin) and cosine (cos), then we have the formula for any rotation (cos&thetax is read as "cosine of the angle times x"):
(cos&theta(x) – sin&theta(y), sin&theta(x) + cos&theta(y))
But that's getting ahead of ourselves. Let's wrap up this basic series by finding the formula for a 75° rotation:
Since we know the formula for any 45° rotation and any 30° rotation, we can just substitute the "x"s and "y"s of one formula for the x's and y's of the other formula to find the formula for a 75° rotation. Then, as we saw before, we can just flip around the coefficients to find the complement's rotation (15°):
To recap, we have derived formulas for 90° (180°, 270°), 45°, and 30° rotations (Click on the degree measures to go to the relevant post. Click on the pics to enlarge):
90°:
(-y, x)45°:
30°:
The pattern to note above--the one I asked about in the previous post--is that the hypotenuses of the triangles can be found in the denominators in the formulas, and the other side lengths of the triangles can be found in the coefficients of x and y in the formulas.
In fact, for both 30° and 45° angles, the rotation formulas can be written as follows:
where, for the time being, opposite side means "the side that is directly opposite the rotation angle in question" and adjacent side means "the side that is not the hypotenuse or the opposite side."
Though we don't really know it yet, this is the formula for any rotation about the origin. That is, if we replace the coefficients above with their somewhat more abstract and useful equivalents, sine (sin) and cosine (cos), then we have the formula for any rotation (cos&thetax is read as "cosine of the angle times x"):
But that's getting ahead of ourselves. Let's wrap up this basic series by finding the formula for a 75° rotation:
Since we know the formula for any 45° rotation and any 30° rotation, we can just substitute the "x"s and "y"s of one formula for the x's and y's of the other formula to find the formula for a 75° rotation. Then, as we saw before, we can just flip around the coefficients to find the complement's rotation (15°):
Labels: mathematics
Old Fraction Division Art
A Pattern Emerges
So, there seems to be a nifty little pattern emerging here. In fact, it's so nifty that it's not only a pattern. It's the pattern we're looking for:
Can you spot it? You can catch up here, here, and then here.
Can you spot it? You can catch up here, here, and then here.
Labels: mathematics
30-Degree Rotations, Part II
Okay. So now that we've got the right perspective, we can work on a formula for 30° rotations about the origin.
Let's take a look very quickly at two of the important representations we used to determine the formula for a 45° rotation about the origin:
From our 45-45-90 triangle, we were able to construct a function table, which took our diagonal side length as an input and returned a horizontal and vertical location (the height and width of the triangle, which had the same length) as an output.
And now that we have our 30-60-90 triangle, we can construct a similar function table.
Once we know the length of the diagonal segment (the length of the x-arm of our "claw"), we can simply divide by 2 to find the vertical location (the y-coordinate) of the end of the arm. To find the horizontal location (the x-coordinate) of the end of the arm, we multiply by the square root of 3 and divide by 2.
So, once again, we take our point at (3, 2) attached to our "claw" and rotate it--this time 30°--about the origin. Using our function table, we can determine the coordinates of the end of the x-arm:
When we turn our attention to the y-arm of our claw, we run into a bit of a snag: the y-arm actually goes up and to the left at a 60° angle, not a 30° angle.
When we rotated our point 45°, we could use the same function table to convert the x-arm length and the y-arm length to coordinate points, because both the x-arm and the y-arm were oriented at 45° angles to the grid. Now it seems that we have to come up with another function table--one that deals with 60° angles--to work with the y-arm of this claw.
Ah, but that is not the case! The triangle we are working with now has a 60° angle. All we need to do to get the right perspective here is to flip this triangle around so that the 60° angle is sitting at the origin:
Using this information and the process we used before, we can determine the final location of the rotated point:
And, of course, the formula for any 30° rotation about the origin:
Let's take a look very quickly at two of the important representations we used to determine the formula for a 45° rotation about the origin:
From our 45-45-90 triangle, we were able to construct a function table, which took our diagonal side length as an input and returned a horizontal and vertical location (the height and width of the triangle, which had the same length) as an output.
And now that we have our 30-60-90 triangle, we can construct a similar function table.
Once we know the length of the diagonal segment (the length of the x-arm of our "claw"), we can simply divide by 2 to find the vertical location (the y-coordinate) of the end of the arm. To find the horizontal location (the x-coordinate) of the end of the arm, we multiply by the square root of 3 and divide by 2.
So, once again, we take our point at (3, 2) attached to our "claw" and rotate it--this time 30°--about the origin. Using our function table, we can determine the coordinates of the end of the x-arm:
When we turn our attention to the y-arm of our claw, we run into a bit of a snag: the y-arm actually goes up and to the left at a 60° angle, not a 30° angle.
When we rotated our point 45°, we could use the same function table to convert the x-arm length and the y-arm length to coordinate points, because both the x-arm and the y-arm were oriented at 45° angles to the grid. Now it seems that we have to come up with another function table--one that deals with 60° angles--to work with the y-arm of this claw.
Ah, but that is not the case! The triangle we are working with now has a 60° angle. All we need to do to get the right perspective here is to flip this triangle around so that the 60° angle is sitting at the origin:
Using this information and the process we used before, we can determine the final location of the rotated point:
And, of course, the formula for any 30° rotation about the origin:
Labels: mathematics
30-Degree Rotations, Part I
So, back to rotations. Last time, we found a formula for determining the coordinates of the image of any point on the coordinate plane rotated 45° (counterclockwise) about the origin. That formula is
where x and y are the coordinates of the original point.
What helped us out tremendously in coming up with this formula was the fact that there is a functional relationship between the length of a line segment oriented at a 45° angle and the horizontal or vertical location of one of its endpoints on the coordinate grid. Once we know the length of any 45° diagonal line segment, in order to find the horizontal or vertical location of a specific endpoint, we can simply divide the length by the square root of 2.
In other words, we were able to come up with a formula because of this triangle . . .
It is natural to wonder whether we can, likewise, identify a constant ratio (a slope) for 30° angles. And of course, with 30° angles, it gets trickier:
As you can see, a line oriented at a 30° angle does not hit gridline intersections as neatly as does the 45° angle line. But we can get some sense of the slope we are looking for by noting where this line does cross intersections. For instance, it comes very close to crossing (14, 8), which gives us a slope of 8/14, or about 0.5714. The point (24, 14) is not as good, but it is close, and it gives us a slope of 0.58333... And (38, 22) seems almost spot-on. At that point, we see a slope of about 0.5789—a number that is, in fact, only about 15 ten-thousandths off from the number we will soon figure out.
As accurate as this sloppy method is, however, it doesn't help us figure out the numbers (the vertical and horizontal side lengths) we can use every time to arrive at it. And, of course, without my having said so earlier, we shouldn't know at this point how accurate this ratio is anyway.
So, how can we figure out exactly what these side lengths should be every time? Let's take a look first at how we might have done it for the 45° angle.
We could have started by noticing that what we were dealing with was a square, whose side lengths are all equal, divided into two congruent triangles by its diagonal. So, we could narrow the three variables of the Pythagorean equation down to two:
The Pythagorean equation then gives us b2 + b2 = c2, or 2b2 = c2.
Solving for c, we get c = b × sqrt(2).
Plugging in that all-important number 1 for b (because we like to think of the axes on the coordinate grid as being labeled in intervals of 1), we get that 45-45-90 triangle mentioned above:
For 30° angles, let's start by taking an equilateral triangle--a triangle with three congruent sides and three congruent angles (each 60°)--turning it on its side by rotating it 90° counterclockwise, and drawing a line through its center as shown below.
The line through the center divides the "top" 60° angle into two 30° angles, and it also divides the base into two congruent segments, each of which is half the length of each of the original sides. If we ignore that bottom right triangle and call these new half-side-lengths b, then we have
Once again, we've narrowed it down to two variables instead of three. Now we just need to figure out what a is in terms of b:
And once again, we can plug in the all-important number 1 for b to get our side lengths:
This means that the slope for any 30° angle will be the same ratio as 1/sqrt(3), or 0.577350269 . . ., which you'll recall is very close to any of the ratios we were getting when we determined this slope from the grid. But now it's exact!
where x and y are the coordinates of the original point.
What helped us out tremendously in coming up with this formula was the fact that there is a functional relationship between the length of a line segment oriented at a 45° angle and the horizontal or vertical location of one of its endpoints on the coordinate grid. Once we know the length of any 45° diagonal line segment, in order to find the horizontal or vertical location of a specific endpoint, we can simply divide the length by the square root of 2.
In other words, we were able to come up with a formula because of this triangle . . .
It is natural to wonder whether we can, likewise, identify a constant ratio (a slope) for 30° angles. And of course, with 30° angles, it gets trickier:
As you can see, a line oriented at a 30° angle does not hit gridline intersections as neatly as does the 45° angle line. But we can get some sense of the slope we are looking for by noting where this line does cross intersections. For instance, it comes very close to crossing (14, 8), which gives us a slope of 8/14, or about 0.5714. The point (24, 14) is not as good, but it is close, and it gives us a slope of 0.58333... And (38, 22) seems almost spot-on. At that point, we see a slope of about 0.5789—a number that is, in fact, only about 15 ten-thousandths off from the number we will soon figure out.
As accurate as this sloppy method is, however, it doesn't help us figure out the numbers (the vertical and horizontal side lengths) we can use every time to arrive at it. And, of course, without my having said so earlier, we shouldn't know at this point how accurate this ratio is anyway.
So, how can we figure out exactly what these side lengths should be every time? Let's take a look first at how we might have done it for the 45° angle.
We could have started by noticing that what we were dealing with was a square, whose side lengths are all equal, divided into two congruent triangles by its diagonal. So, we could narrow the three variables of the Pythagorean equation down to two:
The Pythagorean equation then gives us b2 + b2 = c2, or 2b2 = c2.
Solving for c, we get c = b × sqrt(2).
Plugging in that all-important number 1 for b (because we like to think of the axes on the coordinate grid as being labeled in intervals of 1), we get that 45-45-90 triangle mentioned above:
For 30° angles, let's start by taking an equilateral triangle--a triangle with three congruent sides and three congruent angles (each 60°)--turning it on its side by rotating it 90° counterclockwise, and drawing a line through its center as shown below.
The line through the center divides the "top" 60° angle into two 30° angles, and it also divides the base into two congruent segments, each of which is half the length of each of the original sides. If we ignore that bottom right triangle and call these new half-side-lengths b, then we have
Once again, we've narrowed it down to two variables instead of three. Now we just need to figure out what a is in terms of b:
And once again, we can plug in the all-important number 1 for b to get our side lengths:
This means that the slope for any 30° angle will be the same ratio as 1/sqrt(3), or 0.577350269 . . ., which you'll recall is very close to any of the ratios we were getting when we determined this slope from the grid. But now it's exact!
Labels: mathematics