A few days ago I posted a “Friendly Formula” for the Midpoint Formula.
Today I am presenting a Friendly Formula for the Distance Formula, an important formula in Algebra 1 courses.
First I’m going to present the Friendly Formula for the Distance Formula and demonstrate how to use it. Then I’ll explain why it makes sense.
Buckle your seatbelts ’cause here it is: the distance between any two points on the coordinate plane is simply the SQUARE ROOT of … (the x-distance squared) plus (the y-distance squared).
And here’s an example of how easy it can be to use this formula. Suppose you want the distance between the points (2, 5) and (4, 9).
First figure out how the distance between the x-coordinates, 2 and 4. Well, 4 – 2 = 2, so the x-distance = 2. Now square that x-distance: 2 squared = 4
Next find the distance between the y-coordinates, 5 and 9: Well, 9 – 5 = 4, so the y-distance = 4. Now square that y-distance: 4 squared = 16
Next add the two squared values you just got: 4 + 16 = 20
Finally take the square root of that sum: square root of 20 = root 20.
That final value, root 20, is the distance between the two points.
Now we get to the question of WHY this Friendly Formula makes sense. I will explain that in my next post.
HINT: The Distance Formula is based on the Pythagorean Theorem. See if you can spot the connection.
EXTRA HINT: Make a coordinate plane. Plot the two points I used in this example, and construct a right triangle in which the line connecting these two points is the hypotenuse. If you can figure this out, the “Aha!” moment is a glorious event!
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In algebra we have many formulas to learn. But one problem is that those formulas are often hard to memorize. They are written with variables, and the variables frequently have subscripts, and the truth is that a lot of us don’t really understand what the formulas are saying or how they work. So of course that makes formulas difficult to memorize.
Enter the concept of “friendly formulas.” Friendly formulas are the very same formulas but written in a way that you can understand and therefore memorize much more easily. It’s an idea I have come up with through my many years of algebra tutoring, and idea is included in my Algebra Survival Guide, available through Amazon.com
In this post I describe the “friendly formula” for the midpoint formula.
So as a refresher, what is the midpoint formula all about?
Basically, it lets you find the midpoint of any line segment on the coordinate plane. Think of it this way. There’s some line segment on the coordinate plane called segment AB. That means that it has an endpoint at point A, another at point B. We are given the coordinate of points A and B. We want to find the coordinates of the point right in the middle of points A and B.
Now let’s make this idea easy. Suppose we focus only on the x-coordinates. Suppose the x-coordinate of point A is 2, and the x-coordinate of point B is 6. Ask yourself: what x-coordinate is perfectly in the middle of coordinates 2 and 6? It’s just like asking: what number is right in the middle of 2 and 6 on the number line? Well, wouldn’t that be 4, since 4 is two more than 2 and two less than 6? And indeed it is 4.
But notice that there’s another way to get 4, given the coordinates 2 and 6. We also could have just added 2 and 6 to get 8, and then divided 8 by 2, since 8 ÷ 2 = 4. In other words, we could have TAKEN the AVERAGE of the two x-coordinates, since taking an average of two numbers is adding them and dividing by two.
Could the midpoint formula actually be as easy as taking averages?!
Before we say yes, let’s test this idea for more complicated situations. We just saw that it works when both coordinates are positive. But suppose one coordinate is positive, the other negative. Let’s let one coordinate be – 2, while the other is + 4. What number is right between those two coordinates on the number line? Well, the numbers are 6 apart, right? And half of 6 is 3, so we could just add 3 to – 2, and get + 1 as the point in between them. And we see that + 1 is three away from both – 2 and 4. But could we also get + 1 by averaging -2 and 4? Let’s try: (- 2 + 4) / 2 = 2 / 2 = + 1. Averaging works again.
And finally, what about the case where both coordinates are negative? Suppose one coordinate is – 2, the other – 8. What number is right between those two numbers on the number line? Well, these numbers are also 6 apart, right? And half of 6 is 3, so we could just add 3 to – 8, and get – 5 as the middle. And we see that – 5 is three away from both – 8 and – 2. But can we also get – 5 by averaging – 8 and – 2? Let’s try: (- 8 + – 2) / 2 = – 10 / 2 = -5. Averaging worked here too!
Since the averaging process works for all three cases, this approach does works always, and in fact it is how the midpoint formula works.
The midpoint formula basically just averages the x-coordinates to get the x-coordinate of the midpoint. Then it averages the y-coordinates to get the y-coordinate of the midpoint.
So here is the “friendly formula” for the midpoint of any segment on the coordinate plane: Given a segment whose x- and y-coordinates are known,
MIDPOINT = (AVERAGE of x-coordinates, AVERAGEof y-coordinates)
Of course there’s a standard way to find the y-intercept of any line, and there’s nothing wrong with using that approach.
But the method I’ll present here is a bit faster and therefore easer. And hey, if we can save time when doing math, it’s worth it … right?
So first let’s recall that the y-intercept of any function is the y-value of the function when the x-value = 0. That’s because the y-intercept is the y-value where the function crosses or touches the old, vertical y-axis, and of course all along the y-axis the x-value is always 0 (zero).
So the standard slope-intercept formula is y = mx + b. In a problem asking for the y-intercept, you’ll be given one point that the line passes through (that point’s coordinates will provide you with an x-value and a y-value), and you will also be told the slope of the line (the line’s m-value). So then, to get the b-value, which is the value of the y-intercept, you just grab your y = mx + b equation (dust it off if you haven’t used it in a while), and plug in the three value you’ve been given: those for x, y and m. Then you solve the equation for the one variable that’s left: b, the value of the y-intercept.
Let’s look at an example: a line with a slope of 2 passes through the point (3, 10). What is this line’s y-intercept.
Now, according to the problem, the m-value = 2, the x-value = 3, and the y=value = 10. We just take these values and plug them into the equation: y = mx + b, like this:
10 = (2)(3) + b
After doing these plug-ins, you just solve the equation for b, finding that b = 4. That means that the y-intercept of the line = 4.
Now let’s see how you can do the same problem, but a little bit faster. To do so, we first need to play around with the y = mx + b equation by subtracting the mx-term from both sides, like this:
y = mx + b [Standard equation.] – mx = – mx [Subtracting mx from both sides.] y – mx = b [Result after subtracting.] b = y – mx [Result after flipping left & right sides of the equation above.]
Aha! Look at that final, beautiful equation. This equation has b isolated on the left-hand side. So now if we want to solve for b, all we do is plug in the x, y and m values into the right-hand side of the equation and simplify the value, and the value we get will be the b-value.
For the problem we just solved, with x = 3, y = 10, m = 2, watch how easy it is to solve:
b = y – mx b = 10 – (2)(3) b = 10 – 6 b = 4
So notice that this technique, just like the first technique, reveals that the y-intercept of the line is 4, or (0, 4). The techniques agree, they just get to the same end in slightly different ways.
Notice that with the second, quicker technique, you don’t need to add or subtract any terms. And that’s a key reason that this technique is faster and easier to use than the standard method. So try it out and stick with it if you like it.
Question related to percents of increase and decrease … we know what happens if you increase an original quantity by adding x% of it, then subtracting that same x%: you’ll wind up with the quantity you started with. That is to say, the value of the original quantity will stay the same.
Now let’s pose the same question with regard to multiplication and division. Suppose you take an original quantity and increase it by x%. Then you turn around and decrease the new quantity by x%?
Will the new amount be: a) the same? b) greater? c) less? or d) might it depend on the value of x? What if you reverse the order … you first decrease the quantity by x%, and then you turn around and increase that new quantity by x%. Will the amount you wind up with be the same or different than what you got the other way around?
Many students intuitively think the final result will be the same as the original quantity. That makes this concept fun to teach; math lessons always are more fun when the results go against intuition.
We’ve already said that the amount of the increase and decrease is x%.
Now to keep the math as simple as possible, we’ll just use 1.00 as the original quantity. That way all values that we wind up with will automatically be in percent form.
When we increase our original quantity, 1.00, by x percent, we multiply it by 1.00 + x% = 1.00 + x/100 = 100/100 + x/100 = (100 + x)/100.
Similarly, when we decrease the original quantity, 1.00, by x percent, we multiply it by 1.00 – x% = 1.00 – x/100 = 100/100 – x/100 =(100 – x)/100.
So, then what are the steps performed in the problem? In the first way described, we 1st) increase the original quantity by x%, then 2nd) starting with the increased value, decrease it by x%. Let’s do that now.
We can draw a number of conclusions from this final algebraic expression, boldfaced.
1) It represents the quantity we are left with after increasing, then decreasing a quantity of 1.00, by x%.
2) Since this was done in a general way, this expression serves as a formula to predict the new value for any situation where an original quantity gets increased or decreased by the same percent.
Example: if $2,000 gets increased by 3%, then decreased by 3%, the amount left would be given by this: 2000 [times] (10,000 – 3^2) / 10,000 = 2000(10,000 – 9)/10,000 = 2000(9991/10,000) = 1998.2
3) Since this expression by which the original value gets multiplied does not changed, no matter the original value, no matter the percent, it behooves us to deeply understand this expression: (10,000 – x^2) / 10,000
4) First, let’s notice that this expression does not change whether we increase the original value first, then decrease — or if we switch the order by decreasing the original value first, then increasing the result. We can see this is true because the numerator of the expression: (10,000 – x^2) arises as the product of (100 + x) and (100 – x). The commutative property guarantees that the order of the factors does not change their product.
So the point here is that it doesn’t matter whether we first increase the original quantity, then decrease it, or if we first decrease the original quantity, then increase it; we’ll get the same result each time.
5) Now let’s look at the relationship between the value of the percent, x, and the outcome, the change in the original quantity.
From the expression we see that as x^2 gets larger, the numerator, 10,000 – x^2, will get smaller. And of course, the value of x^2 increases as the value of x increases.
So in a transitive way of thinking, as the percent, x, that we increase and decrease by gets larger, the ultimate decrease of the original value gets larger. Or, slightly differently, as the percent of change we put the value through increases, the original quantity will end up decreasing more at the end.
Here’s a table that shows the relationship between the x-value, the percent of increase and decrease, and the percent of the original value you’re left with after the increase and decrease are carried out.
This is the first time I’ve explored this topic. So feel free to share your thoughts, insights on it.
There are many ways to teach about linear equations, right. You can (a) use the old t-table approach; (b) you can draw a line and then figure out its slope and y-intercept; or you can (c) first explain the slope-intercept formula and then explain how the equation aligns with the formula.
However I would contend that all three of those approaches leaves students somewhat baffled.
The (a) t-table approach makes it appear that points just appear because there’s some machine (the equation) that spits out points in a rather random way, and then these points (the ones on the lines) are just special points that we have found, and perhaps the only ones that are on the line.
The (b) technique gives the impression that — I don’t know —every line just has these two weird properties, slope and y-intercept, and you are simply finding them for this line. While this is true, it does not feel inherently interesting to students, in my experience, partly because slope and y-intercept seem so different from each other. Why would a line necessarily have these two properties? There’s nothing obvious about why a line would have these two properties and no other properties.
The (c) technique is little better than the (b) technique. Students are still left wondering why a line would have these two properties: slope and y-intercept. The two properties seem unrelated to each other and not necessarily essential to the nature of a line.
So is any approach better? Here’s an idea. Get kids thinking about lines by thinking about the relationship between the x- and y-values of the points on the lines.
Start out with this line: y = x. Point out that y = x is in a sense an equation that sends out a call, much as a town squire might shout out a call in a village square for certain people to assemble. While the town squire might be calling for certain men (or women) over a certain age to assemble, the equation y = x, on the other hand, sends out a call for certain pairs of points to assemble, specifically, the pairs of points whose y-values are equal to their x-values, since y = x. Ask students what kind of points those would be. So if the x-value of such a point is 3, what would its y-value have to be? Answer: it would have to be the y-value that is the same as the x-value, so the y-value would also need to be 3. So the point called to assemble would be this point: (3, 3).
Similarly, if x = -3, the associated y-value would be what? Answer: -3. So this point called to assemble would be: (-3, -3). And if x = 0, the associated y-value would be what? Answer: 0. So this point would be: (0, 0). So when these three points are assembled, you would have these points standing bravely on the cold, coordinate plane: (3, 3), (-3, -3), (0, 0). Tell students to graph these three points. When they do so carefully, using a ruler, ask them to name some other points that appear to stand on this same line. They should report that this line also contains points such as (–5, -5), (6, 6), (9 9), (5,5), even (0.5, 0.5), etc.
Students should gradually come to see that when they graph y = x, they wind up graphing the infinitely many points whose y-values are exactly equal to their x-values. And thus they should see that this is NOT a coincidence. They should make the connection that the line, y = x, calls forth the ordered pairs whose y-values are equal to their x-values! This insight is the fundamental concept behind this lesson.
What is so amazing is how many students do NOT realize this most fundamental fact. I have been tutoring students in algebra for 25 years, and I find that when I explain this rather fundamental fact to students, 9 times out of 10 students are amazed.
So I cannot state how important it is that we help students make this connection: there is a relationship between an equation and the coordinate points that the equation generates.
Once we help students see this for y = x, we should move forward in small steps. I will show how I do this in the next post.
Sierra (not her real name) lay down her phone on the tutoring desk and shot me a quick glance, eyes wide.
“Two minutes ago I would’ve had no way to do these in my head,” she said.
Sierra had just completed a round of mental math addition problems through a phone app*. She was adding two two-digit numbers and had just gotten four problems correct in a row.
The app just showed Sierra how to do addition from left to right, not right to left, thus “chunking” the problem.
For example, take the problem 58 + 79. The app encouraged Sierra to add the tens digits first, helping her see that the 5 and 7, respectively, actually represent 50 and 70, which sum to 120.
The app instructed Sierra to hold that 120 in her short-term memory, then sum the digits in the ones place, 8 and 9, to get 17.
Finally, the app told Sierra to add the two subtotals to find the answer: 120 + 17 = 137.
Sierra said she had never realized that an addition problem could be tackled this way, even though she’s now 16 years old. She left the tutoring session saying she was excited to continue working with the app. The best feature of the app, she said, is that it provides tips on how to do each problem quickly and efficiently.
As a longtime tutor, I’ve long noticed that teaching mental math strategies often helps boost students’ “math self-esteem.” Today’s session with Sierra reinforced that point in a big way.
Before today’s session, Sierra had been voicing a range of negative feelings about her relationship to math. She said she can’t be “a good logical thinker” because she struggles with math. She said that she has “always” been bad at math. When I explored that statement, I found that its roots trace back to 4th grade when she had a teacher who forced students to do problems on the board, whether or not they were prepared. On a number of occasions Sierra was called to the board and confused. Rather than helping Sierra through her confusion, the teacher allowed her to wallow in it publicly. I know … terrible, right?!
But, back to the healing side of things … how does the development of mental math skills make such a big difference for many students? I think the answer relates largely to two concepts: labor/efficiency, and power.
If a student lacks good mental math skills, s/he will find the process of doing arithmetic laborious, i.e., like drudgery. If you doubt that, remind yourself of what it felt like to do a long division problem of the sort: 25,682 ÷ 479. By contrast, a student with strong mental math skills finds the process of arithmetic easy and somewhat fun. Knowing those shortcuts is actually kind of exhilarating, especially if you know A LOT of shortcuts.
Mental math skills also make a student feel powerful. For example, a student who has the rules of divisibility solidly locked down will have no problem when learning the skill of prime factorizing numbers; those divisibility rules will make the factoring process fast and fun. On the other hand, a student who doesn’t know divisibility rules will need to rely on multiplication facts (a rather backward way of working), or on a calculator (slow for this kind of task) — just to perform the division aspect of prime factorizing. As a result, this student will not enjoy prime factorizing numbers, and won’t even know how it might be possible to enjoy the process.
So it’s critically important that we help students nurture and maintain a solid toolbox of mental math skills The internet is loaded with resources for mental math. Just go to YouTube and search for this topic. This blog, also, has many articles and videos on the topic. Just search it for mental math, and see what pops up.
* the phone app Sierra was using was Magoosh’s Mental Math Flashcards app.
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