The Midpoint Formula

The Midpoint Formula 

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Sometimes you need to find the point that is exactly between two other points. For instance, you might need to find a line that bisects (divides into equal halves) a given line segment. This middle point is called the "midpoint". The concept doesn't come up often, but the Formula is quite simple and obvious, so you should easily be able to remember it for later. Think about it this way: If you are given two numbers, you can find the number exactly between them by averaging them, by adding them together and dividing by two. For example, the number exactly halfway between 5 and 10 is  ^{[5 + 10]}/₂  =  ¹⁵/₂   = 7.5.
The Midpoint Formula works exactly the same way. If you need to find the point that is exactly halfway between two given points, just average the x-values and the y-values.

• Find the midpoint between (–1, 2) and (3, –6).

Apply the Midpoint Formula:

So the answer is P = (1, –2).

Technically, the Midpoint Formula is the following:

But as long as you remember that you're averaging the two points' x- and y-values, you'll do fine. It won't matter which point you pick to be the "first" point you plug in.

• Find the midpoint between (6.4, 3) and (–10.7, 4).

Apply the Midpoint Formula:

So the answer is P = (–2.15, 3.5)

• Find the value of p so that (–2, 2.5) is the midpoint between (p, 2) and (–1, 3).

I'll apply the Midpoint Formula:   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

This reduces to needing to figure out what p is, in order to make the x-values work:

So the answer is p = –3.

• Is y = 2x – 4.9 a bisector of the line segment with endpoints at (–1.8, 3.9) and (8.2, –1.1)?

If I just graph this, it's going to look like the answer is "yes". But I have to remember that a picture can suggest an answer, it can give me an idea of what is going on, but only the algebra can give me an exact answer. So I'll need to find the midpoint, and then see if the midpoint is actually a point on the given line. First, I'll apply the Midpoint Formula:

Now I'll check to see if this point is on the line:

y = 2x – 4.9 y = 2(3.2) – 4.9 = 6.4 – 4.9 = 1.5
But I needed y to equal 1.4, so this line is close to being a bisector (as a picture would indicate), but it is not exactly a bisector (as the algebra proves). So the answer is "No, this is not a bisector."

• Find the perpendicular bisector of the line segment with endpoints at (–1.8, 3.9) and (8.2, –1.1).   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

This is a multi-part problem, and is actually typical of problems you will probably encounter at some point when you're learning about straight lines. This is an example of a question where you'll be expected to remember the Midpoint Formula from however long ago you last saw it in class. Don't be surprised if you see this kind of question on a test! Here's how to answer it:

First, I need to find the midpoint, since any bisector, perpendicular or otherwise, must pass through the midpoint. I'll apply the Midpoint Formula:

Now I need to find the slope of the line segment. I need this slope in order to find the perpendicular slope. I'll apply the Slope Formula:

The perpendicular slope (for my perpendicular bisector) is the negative reciprocal of the slope of the line segment. Remember that "negative reciprocal" means "flip it, and change the sign". Then the slope of the perpendicular bisector will be + ²/₁ = 2. With the slope and a point (the midpoint, in this case), I can find the equation of the line:

y – 1.4 = 2(x – 3.2) y – 1.4 = 2x – 6.4 y = 2x – 6.4 + 1.4 y = 2x – 5

• Find the center of the circle with a diameter having endpoints at (–4, 3) and (0, 2).

Since the center is at the midpoint of any diameter, I need to find the midpoint of the two given endpoints:

( ^{[–4 + 0]}/₂ , ^{[3 + 2]}/₂ ) = ( ^–4/₂ , ⁵/₂ ) = (–2, 2.5)

These examples really are fairly typical. You will have some simple "plug-n-chug" problems when the concept is first introduced, and then later, out of the blue, they'll hit you with the concept again, except it will be buried in some other type of problem. I'm telling you this now, so you'll know to remember the Formula for later.

The Distance Formula

The Distance Formula 

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The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. Here's how we get from the one to the other:

Suppose you're given the two points (–2, 1) and (1, 5), and they want you to find out how far apart they are. The points look like this:
You can draw in the lines that form a right-angled triangle, using these points as two of the corners:
It's easy to find the lengths of the horizontal and vertical sides of the right triangle: just subtract the x-values and the y-values:

Then use the Pythagorean Theorem to find the length of the third side (which is the hypotenuse of the right triangle):

c² = a² + b²

...so:   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

This format always holds true. Given two points, you can always plot them, draw the right triangle, and then find the length of the hypotenuse. The length of the hypotenuse is the distance between the two points. Since this format always works, it can be turned into a formula:

Distance Formula: Given the two points (x₁, y₁) and (x₂, y₂), the distance between these points is given by the formula:

Don't let the subscripts scare you. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Whichever one you call "first" or "second" is up to you. The distance will be the same, regardless.

• Find the distance between the points (–2, –3) and (–4, 4).

I just plug the coordinates into the Distance Formula:

Then the distance is sqrt(53), or about 7.28, rounded to two decimal places.
The most common mistake made when using the Formula is to accidentally mismatch the x-values and y-values. Be careful you don't subtract an x from a y, or vice versa; make sure you've paired the numbers properly.
Also, don't get careless with the square-root symbol. If you get in the habit of omitting the square root and then "remembering" to put it back in when you check your answers in the back of the book, then you'll forget the square root on the test, and you'll miss easy points.
You also don't want to be careless with the squaring inside the Formula. Remember that you simplify inside the parentheses before you square, not after, and remember that the square is on everything inside the parentheses, including the minus sign, so the square of a negative is a positive.
By the way, it is almost always better to leave the answer in "exact" form (the square root "" above). Rounding is usually reserved for the last step of word problems. If you're not sure which format is preferred, do both, like this:

Now YOU try it!

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Very often you will encounter the Distance Formula in veiled forms. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. For instance:

• Find the radius of a circle, given that the center is at (2, –3) and the point (–1, –2) lies on the circle.

The radius is the distance between the center and any point on the circle, so I need to find the distance: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

Then the radius is sqrt(10), or about 3.16, rounded to two decimal places.

• Find all points (4, y) that are 10 units from the point (–2, –1).
I'll plug the two points and the distance into the Distance Formula:

Now I'll square both sides, so I can get to the variable:

This means y = –9 or y = 7, so the two points are (4, –9) and (4, 7).

If you're not sure why there are two points that solve this exercise, try drawing the (–2, –1) and then drawing a circle with radius 10 around this. Then draw the vertical line through x = 4. You'll see that the vertical line crosses the circle in two spots: (4, –9) and (4, 7).

Slope of a line

Slope of a Straight Line

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One of the most important properties of a straight line is in how it angles away from the horizontal. This concept is reflected in something called the "slope" of the line. Let's take a look at the straight line y = ( ²/₃ ) x – 4. Its graph looks like this:

To find the slope, we will need two points from the line.
Pick two x's and solve for each corresponding y: If, say, x = 3, then y = ( ²/₃ )(3) – 4 = 2 – 4 = –2. If, say, x = 9, then y = ( ²/₃ )(9) – 4 = 6 – 4 = 2. (By the way, I picked the x-values to be multiples of three because of the fraction. It's not a rule that you have to do that, but it's a helpful technique.) So the two points (3, –2) and (9, 2) are on the line y = ( ²/₃ )x – 4.
To find the slope, you use the following formula:

(Why "m" for "slope", rather than, say, "s"? The official answer is: Nobody knows.)
The subscripts merely indicate that you have a "first" point (whose coordinates are subscripted with a "1") and a "second" point (whose coordinates are subscripted with a "2"); that is, the subscripts indicate nothing more than the fact that you have two points to work with. It is entirely up to you which point you label as "first" and which you label as "second". For computing slopes with the slope formula, the important thing is that you subtract the x's and y's in the same order. For our two points, if we choose (3, –2) to be the "first" point, then we get the following:

The first y-value above, the –2, was taken from the point (3, –2) ; the second y-value, the 2, came from the point (9, 2); the x-values 3 and 9 were taken from the two points in the same order. If we had taken the coordinates from the points in the opposite order, the result would have been exactly the same value:

As you can see, the order in which you list the points really doesn't matter, as long as you subtract the x-values in the same order as you subtracted the y-values. Because of this, the slope formula can be written as it is above, or alternatively it can be written as:

Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

Let me emphasize: it does not matter which of the two formulas you use or which point you pick to be "first" and which you pick to be "second". The only thing that matters is that you subtract your x-values in the same order as you had subtracted your y-values.

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Technically, the equivalence of the two slope formulas above can be proved by noting that:

y₁ – y₂ = –y₂ + y₁ = –(y₂ – y₁)
x₁ – x₂ = –x₂ + x₁ = –(x₂ – x₁)

Doing the subtraction in the so-called "wrong" order serves only to create two "minus" signs which cancel out. The upshot: Don't worry too much about which point is the "first" point, because it really doesn't matter. (And please don't send me an e-mail claiming that the order does somehow matter, or that one of the above two formulas is somehow "wrong". If you think I'm wrong, plug pairs of points into both formulas, and try to prove me wrong! And keep on plugging until you "see" that the mathematics is in fact correct.)

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Let's find the slope of another line equation:

• Find the slope of  y = –2x + 3.

Graphing the line, it looks like this:

I'll pick a couple of values for x, and find I'll find the corresponding values for y. Picking x = –1, I get y = –2(–1) + 3 = 2 + 3 = 5. Picking x = 2, I get y = –2(2) + 3 = –4 + 3 = –1. Then the points (–1, 5) and (2, –1) are on the line y = –2x + 3. The slope of the line is then calculated as:

Now YOU try it!

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Scroll back up this page and look at those equations and their graphs. For the first equation, y
= ( ²/₃ )x – 4, the slope was m = ²/₃. And the line, as you moved from left to right along the x-axis, was heading up toward the top of the drawing; technically, the line was "increasing". For the second line, y = –2x + 3, the slope was m = –2. And the line, as you moved from left to right along the x-axis, was heading down toward the bottom of the drawing; technically, the line was "decreasing". This relationship is always true: Increasing lines have positive slopes, and decreasing lines have negative slopes. Always!
This fact can help you check your calculations: if you calculate a slope as being negative, but you can see from the graph that the line is increasing (so the slope must be positive), you know you need to re-do your calculations. Being aware of this connection can save you points on a test because it will enable you to check your work before you hand it in.

Increasing lines have positive slopes; decreasing lines have negative slopes. With this in mind, consider the following horizontal line: y = 4 Its graph is shown to the right.

Is the horizontal line going up; that is, is it an increasing line? No, so its slope won't be positive. Is the horizontal line going down; that is, is it a decreasing line? No, so its slope won't be negative. What number is neither positive nor negative? Zero! So the slope of this horizontal line is zero. Let's do the calculations to confirm this value. Using the points (–3, 4) and (5, 4), the slope is:

This relationship is true for every horizontal line: a slope of zero means the line is horizontal, and a horizontal line means you'll get a slope of zero. (By the way, all horizontal lines are of the form "y = some number", and the equation "y = some number" always graphs as a horizontal line.)

Now consider the vertical line x = 4: Is the vertical line going up on one end? Well, kind of. Is the vertical line going down on the other end? Well, kind of. Is there any number that is both positive and negative? Nope.

Verdict: vertical lines have NO SLOPE. In particular, the concept of slope simply does not work for vertical lines. The slope doesn't exist! Let's do the calculations. I'll use the points (4, 5) and (4, –3); the slope is:

(We can't divide by zero, which is of course why this slope value is "undefined".)
This relationship is always true: a vertical line will have no slope, and "the slope is undefined" means that the line is vertical. (By the way, all vertical lines are of the form "x = some number", and "x = some number" means the line is vertical. Any time your line involves an undefined slope, the line is vertical, and any time the line is vertical, you'll end up dividing by zero if you try to compute the slope.)
Warning: It is very common to confuse these two lines and their slopes, but they are very different. Just as "horizontal" is not at all the same as "vertical", so also "zero slope" is not at all the same as "no slope". The number "zero" exists, so horizontal lines do indeed have a slope. But vertical lines don't have any slope; "slope" just doesn't have any meaning for vertical lines. It is very common for tests to contain questions regarding horizontals and verticals. Don't mix them up!
Slope: Parallel and Perpendicular Lines

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Parallel lines and their slopes are easy. Since slope is a measure of the angle of a line from the horizontal, and since parallel lines must have the same angle, then parallel lines have the same slope — and lines with the same slope are parallel.

Perpendicular lines are a bit more complicated. If you visualize a line with positive slope (so it's an increasing line), then the perpendicular line must have negative slope (because it will be a decreasing line). So perpendicular slopes have opposite signs. The other "opposite" thing with perpendicular slopes is that their values are reciprocals; that is, you take the one slope value, and flip it upside down. Put this together with the sign change, and you get that the slope of the perpendicular line is the "negative reciprocal" of the slope of the original line — and two lines with slopes that are negative reciprocals of each other are perpendicular to each other. In numbers, if the one line's slope is m = ⁴/₅, then the perpendicular line's slope will be m = ^–5/₄. If the one line's slope is m = –2, then the perpendicular line's slope will be m = ¹/₂.
In your homework, you will probably be given some pairs of points, and be asked to state whether the lines through the pairs of points are "parallel, perpendicular, or neither". To answer the question, you'll have to calculate the slopes and compare them. Here's how that works:

• One line passes through the points (–1, –2) and (1, 2); another line passes through the points (–2, 0) and (0, 4). Are these lines parallel, perpendicular, or neither?

To answer this question, I'll find the slopes.

Since these two lines have identical slopes, then these lines are parallel.

• One line passes through the points (0, –4) and (–1, –7); another line passes through the points (3, 0) and (–3, 2). Are these lines parallel, perpendicular, or neither?

I'll find the values of the slopes. Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

 
If I were to flip the "3" and then change its sign, I would get " ^–1/₃". In other words, these slopes are negative reciprocals, so the lines through the points are perpendicular.

• One line passes through the points (–4, 2) and (0, 3); another line passes through the points (–3, –2) and (3, 2). Are these lines parallel, perpendicular, or neither?

I'll find the slopes.

 
These slope values are not the same, so the lines are not parallel. The slope values are not negative reciprocals either, so the lines are not perpendicular. Then the answer is "neither".

Warning: When asked a question of this type ("are they parallel or perpendicular?"), do not start drawing pictures. If the lines are close to being parallel or close to being perpendicular (or if you draw the lines messily), you can very-easily get the wrong answer from your picture. Besides, they're not asking if the lines look parallel or perpendicular; they're asking if the lines actually are parallel or perpendicular. To be sure of your answer, do the algebra.

x- and y-Intercepts

x- and y-Intercepts

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The graphical concept of x- and y-intercepts is pretty simple. The x-intercepts are where the graph crosses the x-axis, and the y-intercepts are where the graph crosses the y-axis. The problems start when we try to deal with intercepts algebraically.
To clarify the algebraic part, think again about the axes. When you were first introduced to the Cartesian plane, you were shown the regular number line from elementary school (the x-axis), and then shown how you could draw a perpendicular number line (the y-axis) through the zero point on the first number line. Take a closer look, and you'll see that the y-axis is also the line "x = 0". In the same way, the x-axis is also the line "y = 0".
Then, algebraically,

• an x-intercept is a point on the graph where y is zero, and
• a y-intercept is a point on the graph where x is zero.

More specifically,

• an x-intercept is a point in the equation where the y-value is zero, and
• a y-intercept is a point in the equation where the x-value is zero.
     

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• Find the x- and y-intercepts of 25x² + 4y² = 9

Using the definitions of the intercepts, I will proceed as follows:
x-intercept(s):

y = 0 for the x-intercept(s), so:

25x² + 4y² = 9
25x² + 4(0)² = 9
25x² + 0 = 9
x² =  ⁹/₂₅
x = ± ( ³/₅ )
Then the x-intercepts are the points ( ³/₅, 0) and ( ^–3/₅, 0)
y-intercept(s):   Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved

x = 0 for the y-intercept(s), so:

25x² + 4y² = 9
25(0)² + 4y² = 9
0 + 4y² = 9
y² =  ⁹/₄
y = ± ( ³/₂ )
Then the y-intercepts are the points (0, ³/₂ ) and (0, ^–3/₂ )

Just remember: Whichever intercept you're looking for, the other variable gets set to zero.

Inequalities

Solving linear inequalities is very similar to solving linear equations, except for one small but important detail: you flip the inequality sign whenever you multiply or divide the inequality by a negative. The easiest way to show this is with some examples:

1)	Graphically, the solution is:	The only difference between the linear equation "x + 3 = 2" and this linear inequality is that I have a "less than" sign, instead of an "equals" sign. The solution method is exactly the same: subtract 3 from either side. Note that the solution to a "less than, but not equal to" inequality is graphed with a parentheses (or else an open dot) at the endpoint, indicating that the endpoint is not included within the solution.
2)	Graphically, the solution is:	The only difference between the linear equation "2 – x = 0" and this linear inequality is the "greater than" sign in place of an "equals" sign. Note that "x" in the solution does not "have" to be on the left. However, it is often easier to picture what the solution means with the variable on the left. Don't be afraid to rearrange things to suit your taste.
3)	Graphically, the solution is:	The only difference between the linear equation "4x + 6 = 3x – 5" and this inequality is the "less than or equal to" sign in place of a plain "equals" sign. The solution method is exactly the same. Note that the solution to a "less than or equal to" inequality is graphed with a square bracket (or else a closed dot) at the endpoint, indicating that the endpoint is included within the solution.
4)	Graphically, the solution is:	The solution method here is to divide both sides by a positive two.     Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved
5)	Graphically, the solution is:	This is the special case noted above.  When I divided by the negative two, I had to flip the inequality sign.

The rule for example 5 above often seems unreasonable to students the first time they see it. But think about inequalities with numbers in there, instead of variables. You know that the number four is larger than the number two: 4 > 2. Multiplying through this inequality by –1, we get –4 < –2, which the number line shows is true:

If we hadn't flipped the inequality, we would have ended up with "–4 > –2", which clearly isn't true.
The previous inequalities are called "linear" inequalities because we are dealing with linear expressions like "x – 2" ("x > 2" is just "x – 2 > 0", before you finished solving it). When we have an inequality with "x²" as the highest-degree term, it is called a "quadratic inequality". The method of solution is more complicated.

• Solve x² – 3x + 2 > 0

First, I have to find the x-intercepts of the associated quadratic, because the intercepts are where y = x² – 3x + 2  is equal to zero. Graphically, an inequality like this is asking me to find where the graph is above or below the x-axis. It is simplest to find where it actually crosses the x-axis, so I'll start there.
Factoring, I get x² – 3x + 2 = (x – 2)
(x – 1) = 0, so x = 1 or x = 2. Then the graph crosses the x-axis at 1 and 2, and the number line is divided into the intervals (negative infinity, 1), (1, 2), and (2, positive infinity). Between the x-intercepts, the graph is either above the axis (and thus positive, or greater than zero), or else below the axis (and thus negative, or less than zero).
There are two different algebraic ways of checking for this positivity or negativity on the intervals. I'll show both.
1) Test-point method. The intervals between the x-intercepts are (negative infinity, 1), (1, 2), and (2, positive infinity). I will pick a point (any point) inside each interval. I will calculate the value of y at that point. Whatever the sign on that value is, that is the sign for that entire interval.
For (negative infinity, 1), let's say I choose x = 0; then y = 0 – 0 + 2 = 2, which is positive. This says that y is positive on the whole interval of (negative infinity, 1), and this interval is thus part of the solution (since I'm looking for a "greater than zero" solution).
For the interval (1, 2), I'll pick, say, x = 1.5; then y = (1.5)² – 3(1.5) + 2 = 2.25 – 4.5 + 2 = 4.25 – 4.5 = –0.25, which is negative. Then y is negative on this entire interval, and this interval is then not part of the solution.
For the interval (2, positive infinity), I'll pick, say, x = 3; then y = (3)² – 3(3) + 2 = 9 – 9 + 2 = 2, which is positive, and this interval is then part of the solution. Then the complete solution for the inequality is x < 1 and x > 2. This solution is stated variously as:


	inequality notation
	interval, or set, notation
	number line with parentheses (brackets are used for closed intervals)
	number line with open dots (closed dots are used for closed intervals)

The particular solution format you use will depend on your text, your teacher, and your taste. Each format is equally valid.   Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved
2) Factor method. Factoring, I get y = x² – 3x + 2 = (x – 2)(x – 1). Now I will consider each of these factors separately.
The factor x – 1 is positive for x > 1; similarly, x – 2 is positive for x > 2. Thinking back to when I first learned about negative numbers, I know that (plus)×(plus) = (plus), (minus)×(minus) = (plus), and (minus)×(plus) = (minus). So, to compute the sign on y = x² – 3x + 2, I only really need to know the signs on the factors. Then I can apply what I know about multiplying negatives.


First, I set up a grid, showing the factors and the number line.
Now I mark the intervals where each factor is positive.
Where the factors aren't positive, they must be negative.
Now I multiply up the columns, to compute the sign of y on each interval.

Then the solution of x² – 3x + 2 > 0 are the two intervals with the "plus" signs:

(negative infinity, 1) and (2, positive infinity).

• Solve –2x² + 5x + 12 < 0.

First I find the zeroes, which are the endpoints of the intervals: y = –2x² + 5x + 12 =
(–2x – 3)(x – 4) = 0 for x = –³/₂ and x = 4. So the endpoints of the intervals will be at –³/₂ and 4. The intervals are between the endpoints, so the intervals are (negative infinity, ^–3/₂], [^–3/₂, 4], and [4, positive infinity). (Note that I use brackets for the endpoints in "or equal to" inequalities, instead of parentheses, because the endpoints will be included in the final solution.)
To find the intervals where y is negative by the Test-Point Method, I just pick a point in each interval. I can use points such as x = –2, x = 0, and x = 5.
To find the intervals where y is negative by the Factor Method, I just solve each factor: –2x – 3 is positive for –2x – 3 > 0, –3 > 2x, –3/2 > x, or x < –³/₂; and x – 4 is positive for x – 4 > 0,
x > 4. Then I fill out the grid:

    

Then the solution to this inequality is all x's in
(negative infinity, ^–3/₂ ] and [4, positive infinity).