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.
Friendly Formulas make algebra less intimidating!
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.
“Friendly” Formulas make it easier to learn and memorize algebraic formulas.
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.
High-Octane Boost for Math Ed
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.
If you’re staring at two terms you need to factor, but feel like a deer looking at the headlights of an oncoming semi, here’s a way to leap to safety!
It’s called the “Difference of Two Squares” trick.
It requires four simple steps.
Figure out if each of the terms is a “perfect square.”
If so, take the square root of each term.
Put each square root in its proper place inside two ( ).
Put a + sign inside the first ( ), and put a – sign inside the second ( ).
Let’s do an easy example. Suppose the terms you’re looking at are these: x^2 – 9
Let’s go through the 4 steps together.
Figure out if each term is a “perfect square.”
So, what does it mean for a number or term to be a “perfect square”? It means that you get the number or term by multiplying a number or term by itself. For example, 16 is a perfect square because you can get 16 by multiplying 4 by itself: 4 x 4 = 16.
So when we look at our two terms, x^2 and 9, we notice that both are perfect squares. 9 is just 3 times 3. And in the same way, x^2 is just x times x.
Take the square root of each term. The square root of x^2 is just x.
And the square root of 9 is just 3.
Put each square root in the proper place inside two sets of ( ). We put the square root of the term that was positive first, and the square root of the term that was negative second.Since the x^2 was the positive term, we put its square root, x, first inside each
( ). So far, that gives us: (x ) (x )
Since the 9 was the negative term because it had the negative sign in front of it: – 9, we put its square root, 3, second inside each ( ). So our ( )s now look like this: (x 3) (x 3)
Finally, we just need to put in signs that connect the terms inside
the ( )s.
That’s easy. We put a + sign inside one ( ), and we put a – sign
inside the other ( ).
I prefer to put the + inside the first ( ), but it really doesn’t matter.The final factored form, then, looks like this: (x + 3) (x – 3)
That’s all there is to it.
Now try these problems for practice.
a) x^2 – 16 b) x^2 – 100 c) x^2 – 121 d) x^4 – 16x^2 e) 49x^8 – 144y^12
Here’s a super-quick shortcut for DIVIDING ANY NUMBER by a RADICAL.
Note: I’m using this symbol (√) to mean square root.
So √5 means the square root of 5; √bmeans the square root of b, etc. And … if you want to learn why this “hack” works, see my explanation at the end of the blog.
This “hack” lets you mentally do problems like the following three. That means you can do these problems in your head rather than on paper.
a) 12 / √3
b) 10 / √2
c) 22 / √5
Here are three terms I’ll use in explaining this “hack.”
In a problem like 12 divided by √3, which I write as: 12 / √3,
12 is the dividend,
3 is the number under the radical,
√3 is the radical.
The “Hack,” Used for 12 / √3:
Divide the dividend by the number under the radical.
In this case, 12 / 3 = 4.
Take the answer, 4, and multiply it by the radical. 4 x √3 = 4√3
Shake your head in amazement because that, right there, is the ANSWER!
Another Example: 10 / √2
Divide the dividend by the number under the radical.
In this case: 10 / 2 = 5
Take the answer you get, 5, and multiply it by the radical. 5 x √2 = 5√2. (Don’t forget to shake head in amazement!)
Third Example: 22 / √5
Divide dividend by number under the radical.
In this case, 22 divided by 5 = 22/5 (Yep, sometimes you wind up with a fraction or a decimal; that’s why I’m giving an example like this.)
Take the answer you get, 22/5, and multiply it by the radical. 22/5 x √5 = 22/5 √5. [Note: the √5 is in the numerator, not
in the denominator. To make the location of this √5 clear, it’s best
to write the answer: 22 √5 / 5].
NOW TRY YOUR HANDby doing these PRACTICE PROBLEMS:
a) 18 / √3
b) 16 / √2
c) 30 / √5
d) 10/√3
e) 12/√5
– – – – – – – – – – – – – – – – – –
ANSWERS:
a) 18 / √3 = 6√3
b) 16 / √2 = 8√2
c) 30 / √5 = 6√5
d) 10 / √3 = 10√3/3
e) 12 / √5 = 12√5/5
– – – – – – – – – – – – – – – – – –
WHY THE “HACK” WORKS:
It works because we rationalize the denominator of a fraction whenever the denominator contains a radical. Here’s the “hack” in general terms, with:
a = the dividend,
b = the number under the radical,
√b = the radical.
a / √b
= a √b
= a √b = a √b √b √b b
Notice: we started with: a / √b.
And keeping things equal, we ended up with a √b / b.
This shows that the “hack” works in general. So it works in all specific cases as well!
– – – – – – – – – – – – – – – – – –
Final note: the number under the radical is called the radicand. But that term is so close to the term radical that I thought it would be less confusing if I just called this the number under the radical. I hope you are not offended.
So, you’d think that combining a positive number and a negative number would be a fairly straightforward thing, huh?
Well, unfortunately, a lot of students think it’s easy. They think it’s too easy. They think there’s one simple rule that guides them to the very same kind of answer every time. And that’s exactly where they get into trouble.
The truth is that combining a positive and a negative number is a fairly complicated operation, and the sign of the answer is dependent on a nmber of factors.
This video reveals a common mistake students make when tackling these problems. it also shows the correct way to approach these problems, using the analogy of having money and owing money to make everything make sense.
So take a look and see if this explanation doesn’t end the confusion once and for all.
And don’t forget: there are practice problems at the end of the video. Do those to make sure you’ve grasped the concept.
Here’s a common mistake, and a very understandable one, too. Students need to combine two negative numbers, and they, of course, wind up with an answer that’s positive. Why? Because, they’ll say — pointing out that you yourself have told them this — “Two negatives make a positive!”
This video gets to the root of this common misunderstanding by helping students understand exactly when two negatives make a positive, and when they don’t.
Make sure you watch the whole video, as there are practice problems at the end, along with their answers.
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