Understanding Point Slope Form

When you are graphing straight lines, one of the most common formats for describing the equation of the line is called "point slope form."  In this representation, the equation identifies one ordered pair that is on the line, and the slope.  If you were given only those pieces of information, you would have all that you would need to construct the line.  Continue reading to learn more about this line graphing concept, and make sure that you Like my post if it is helpful to you!

By using this formula...
- if you know one point that is on the line (x₁, y₁),
- and you know the slope of the line (m),
... with a little bit of mathematics and algebraic rearrangement, you can determine any other point (x, y) on the line.

Here's one type of problem that you will likely encounter:  Express in point slope form the line that passes through (4, 2) and has a slope of 8.  

To correctly solve this problem, all that you need to do is substitute the given values into the equation shown above.  Therefore, the correct expression is y - 2 = 8(x - 4).  It's as easy as that!

A more complicated problem would be something like this: What is the y-coordinate when x = 3 on the line that passes through (1, 1) and has a slope of 5?

To successfully work through this problem, you first approach it as you did the previous one.  That is, find the equation of the line.  In this case it is y - 1 = 5(x - 1).  Now, to find y when x = 3 is as simple as subbing in x = 3 into this equation, doing a bit of rearranging, and simplifying to isolate y.  Try it out for yourself, and you will see that y = 11 when x = 3.  In other words, the point (3, 11) is on the line that is described in the question.

Now that you've seen a few questions that can be asked about point slope form, perhaps it might help you to better understand the concept if you see what it actually means.  First off, consider all of the variables that are included in the expression:

We have an ordered pair (x₁, y₁), an unknown point (x, y) that can be any point on the line at all, and the slope (m).  Now, my question to you is: where have you seen these variables together in one place before?

If you answered that this is a rearrangement of the slope formula, then you get a gold star!  As we've seen before, the slope of a line is equal to rise over run.  In other words, the slope corresponds to the ratio of the change in vertical height to the change in horizontal distance of the line.  Mathematically, here is what you this means:

So, really, all we're really dealing with for any of this is the definition of the slope!  The point slope formula is just a different way of looking at it.

Now, it should be said that this may not be the most intuitive way of representing an equation of a line. Many people find expressing their equation in the form of y = mx + b to be more familiar and descriptive, since by definition it denotes the slope and y-intercept.  The intercept is a very easy "starting point" from which to extend your line, and with the known slope, is it simple to count spaces to plot another point.  Whichever way you express the equation of your line, assuming that the math is correct, they are just different ways of describing the same thing and so they aren't wrong... of course, unless your teacher specifically asks for one form or the other.

If you think about it, you should be able to see the connection between "point slope" and "slope intercept" forms.  Consider that in y = mx + b, the b term is the y-intercept (or y₁), which means that x₁ = 0.  So, you can say that:

y = mx + y₁

y - y₁ = mx

Looks familiar, except it is missing the x₁ term... but we already designated that it is 0 anyways!

Another way to show that these two formulas are the same thing is to equate them to the same variable, which therefore means the equations are equal.  This is like solving a system of equations.

If we solve for m in the point slope equation, we have: m = (y - y₁) / (x - x₁).
If we solve for m in the slope intercept equation, we have: m = (y - b) / x.

Since we're talking about the same line, it obviously is the same slope in each version, so it is fair to equate them. So:

(y - y₁) / (x - x₁) = (y - b) / x

Written like this, it is easy to see how the two equations correspond to each other.

A third way of representing a line's equation is to express it in standard form.  What this does is put everything on to one side of the expression, simplifies it, and sets it equal to 0.  Your expression will then have the form Ax + By + C = 0, where A is the "x" coefficient, B is the "y" coefficient, and C is the constant not associated with either x or y.

The whole point of all of this is to make a very simple point: all you need to fully describe the equation of a straight line is its slope and a point on it.  If you only have two points given, it is easy to calculate the slope from the slope formula, and then it is only a matter of plugging numbers into whichever expression you like.  If you only have the slope and no points, then you have the shape of a line but no point to which you can anchor it.  Get the slope, get a point, pick a way of expressing the equation of your line, and that's all there is to it.  If you made it all the way to the end, please remember to click the Like or +1 buttons (or both) below if you enjoyed this post!  Thanks!

How to Derive the Equation of a Circle

Continuing from my last post about the Equation of a Circle, here I would like go through an exercise which hopefully explains more about "why" the equation looks the way it does.  The equation of a circle isn't a difficult one to memorize, and the modifications that you can make to it to shift it either horizontally or vertically are very similar to the methods used in translating graphs of functions.  However, the theory behind the equation is very distinctively about a circle.

To begin, let's start with a point (5, 7).  Now, I ask to find points that are exactly 5 units away from this point.

To begin, it is very straightforward to deduce points that are horizontally and vertically in line with the center point.  We simply move 5 units up and down, and left and right.  This allows us to identify the points (10, 7), (0, 7), (5, 12), and (5, 2).

After these points, it seems that finding new points gets a bit more difficult.  However, it should be fairly obvious that by asking for points that are 5 units away from the center, this is essentially asking for points along a circle with radius of 5.  So then, how to find another set of coordinates (x, y)?

If you pay attention to how I have purposefully drawn the radius in the above picture, you will notice that I have inscribed a right angle triangle with a hypotenuse (radius) of 5.  This is one of the special triangles, a 3-4-5 triangle.  So then, if we then use the dimensions of this triangle, and rotate and flip it around the circle, we can come up with several more exact points that are on the circle: (8, 11), (9, 10), (8, 3), (9, 4), (2, 11), (1, 10), (2, 3), (1, 4).

As you can see, the points that we have identified are beginning to fill out the circle.  But, how can we identify even more (x, y) points?  It doesn't look like there are any points that lie on whole number integer coordinates left.  All that remains are fractional coordinates, and that seems like a lot more work than it's worth to identify more points!  Maybe we can find an expression to more accurately pinpoint ordered pairs, that will include all of those that we've already identified plus all of the ones that we can't?  Let's go back to our picture, and note the generic triangle now inscribed.

If we say that our radius of 5 is the hypotenuse of a right angle triangle, and we already know what the center point of our circle is, then we can do a bit of arithmetic to come up with generic side lengths for our triangle.  If we sweep the radius around the circle, and we always let the inscribed point be called (x, y), then we can state expressions that denote the length of the sides.  In this example, the horizontal side is equal to x-5.  I've visualized this in the following image.

Similarly, the vertical side length of our new triangle can be shown to be y-7.

Now, since we have described a right angle triangle with defined sides, we can apply the famous Theorem of Pythagoras to this triangle (a² + b² = c²).  This means that we can come up with the following equation:

And by now, you should recognize that this is the form of the equation of a circle:

Quite simply, this equation looks the way it does because it is based on the Pythagoras Theorem.  The circle equation really isn't a difficult one to have to memorize, but hopefully this demonstration has shown you the value of triangles and geometry in deriving more advanced math concepts.  Please don't forget to hit the +1 button below, and also follow me on twitter!  You can even click here to tweet about my post.

Credit to a YouTube video by DrJamesTanton where I got the idea for this post.

Equation of a Circle

Once you have worked with functions for a while, inevitably you will begin to wonder "what about circles?"  You have explored all sorts of different equations and their graphs.  Perhaps you've even gone so far as to rotate graphs sideways and learn how to manipulate those equations as well.  But despite these seemingly more involved concepts, you've yet to come across circles.  For such an apparently simple shape, why have these not been included in such rudimentary graphing lessons.

Well, circles are a little different from what you've done so far.  For starters, circles technically aren't functions.  This may surprise you at first, but recall the vertical line test.  Would a circle pass such a test?  Of course not, because passing a straight vertical line through any point on a circle (except for the tangent points on the sides) would also intersect the circle on its opposite side.  Try it and you will see!  So, if a circle is not a function, then how do we handle describing their equations?  That is a little different, but really not much harder than your typical parabola graph.  Follow along and I will explain the equation of a circle in more detail.

For starters, let me just show you the equation for a simple circle.  Consider a circle with a center at the origin (0, 0) with a radius of 1.  We call this a unit circle.  Here is what the equation looks like:

It looks simple, right?  This equation can also be modified using similar concepts to how you would manipulate a function, such as a parabola.  There is a simple change to make to the equation that causes the graph of the function to translate left or right, and a second similar change you can make that results in a vertical translation of the graph.  In this case, depending on where you want to shift your circle, that is the variable you modify.

So, if you want to center your circle at (2, 0), which is a horizontal translation of 2 units to the right, you would change the above formula to include an (x-2)2 term.  Think of it as "if x=2, the whole x term becomes 0.  So 2 is its root."  Same thing applies to vertical translation.  If you change the y term to be (y-5)2, then you can see that the whole y term becomes 0 when y=5, so this means that the circle is centered at a height of y=5.  If we combine these two translations, we shift the circle to have a center of (2, 5), and the final equation looks like this:

Hopefully you can see how easy it can be to locate the center of a circle based on its equation, and how equally simple it can be to determine the equation of a circle just by a visual inspection.  However, to fully describe a circle, there is still something missing.  We haven't looked at what the "1" means.

To fully describe a circle mathematically, the only things that you need to know about it are the coordinates of its center, as well as its radius.  You may think that you should need more information, but think of it as the mathematical equivalent of a geometry set's compass.  To draw a perfect circle with a compass, all you do is put the point down at the center, set the radius, and spin the pencil around the paper.  You only needed those two pieces of information to be able to construct a circle.

Continuing with our example, the radius of our circle is described by the 1 term.  Technically, you can think of it as a 12 term, which provides for our circle to have a radius of 1 (or a unit circle).  To change our radius to 5, we would change the 1 to a 25, because just like the x and y terms, the radius term is also squared.  If we wanted a radius of 7, we would put the term as 49.  In general, this term is r2.

If we put all of these components together, we can come up with the general form for the equation of a circle.  We can include terms that allow for horizontal and vertical translation, as well as whatever radius we want.  Let's call the horizontal shift term "a" and the vertical shift term "b", and the radius "r".  In this case, here is the general form of the equation of a circle:

The equation describes a perfect circle, and doesn't allow for any stretching or compressing along either of the axes.  All that needs to be done is determine the circle's center point and radius, and you can easily fill in the relevant values.

As an example, consider the circle that has radius of 10, and is centered at (-5, 7).  Here is its equation:

Here is the opposite form of the question: what is the center and radius of the circle described by the following equation:

Quick analysis of this by comparison to the general form allows you to find that the center point is (-12, -2) and has a radius of 12.

Hopefully you can see that graphing circles is a little bit different from graphing more familiar functions, though still similar enough that the math concepts we've learned so far can easily be adapted to apply here as well.  In my next post, I would like to explain a little about just "why" the equation of a circle looks the way it does.  For teachers, this is also a good exercise to have students explore when first learning how to graph circles.  

Remember to please click the +1 button below to share my post!

What is Perpendicularity?

"Perpendicular" is the term used in mathematics to describe two lines that intersect at right angles.  I recently introduced this concept in a separate post about the definition of perpendicular lines, but I thought it might be interesting to go into a little bit more detail about this rudimentary and familiar math concept.  For students who are just learning about graphing lines for the first time, this is undoubtedly sufficient, but for those more familiar with the concept, this post might provide additional insight.

I opened above by stating that we are talking about two intersecting lines that form a 90 degree angle with each other, though this definition can and should be expanded.   Though technically correct, there is more to the concept of "perpendicularity," as it's called when talking about this subject.  More correctly, this term applies to not only lines (and line segments), but planes (surfaces, not airplanes!) as well.  At first this may sound advanced, but if you think about it, it is completely obvious. If you stack two books together in a perpendicular arrangement, you essentially are viewing the intersection of two planes.  Lines are drawn on paper, but planes are like the three-dimensional versions that have depth., and you can readily find examples in real life of things that are at 90 degrees to each other.

That leads me to a second point about perpendicular lines and planes.  I have said that they intersect at 90 degrees, which is true again, but more accurately and mathematically you can say that the lines form two "congruent adjacent angles" (Wikipedia link).  This means that if you look at the perpendicular intersection in a T-shape, you see two angles next to each other (adjacent) that are the same (congruent).  And by the rules of geometry, since the angles along a straight line must add up to 180 degrees, this obviously means that the intersection must be composed of 90 degree angles.  Furthermore, if the lines extend through each other, the rules of geometry state that all angles around a point must sum up to 360 degrees, so again we have 90 degrees for each - four right angles.

Perpendicular lines

Perpendicular planes

Here's some additional information that you might find interesting.  The term "perpendicular" itself can be an adjective describing the lines, as in "the perpendicular lines are written in red ink."   Alternatively, it can be the noun, as in "the perpendicular to the ground rises to the sky."   Sorry, those examples aren't very creative!  Also, another word for perpendicular is "orthogonal," which can be used the same way to describe right angle intersections.  "Orthogonality"comes from the Greek for "straight angle" and refers to lines and planes at ninety degree angles, much like "perpendicularity."

I know these math definitions probably aren't going to make solving your graphing problems any easier, but I thought that it was good information to know!  It's a very important but basic concept that is introduced very early in math education, so hopefully I have explained it simply enough even for beginners to grasp.  Please click the +1 button below to share my post, and you can even tweet about my site if you like it!

Definition of Perpendicular Lines

A lot of traffic coming to my site lately has been specifically seeking to learn more about perpendicular lines.  At this point in the new school year, many math classes are just beginning to study graphing, and so perpendicular and parallel lines are often discussed along with other basic graphing concepts, such as slopes and intercepts.  In this post, I would like to give you a definition of perpendicular lines, maybe include a bit of review of some other graphing notes, and go over a few example questions involving them that you are almost guaranteed to encounter in your mathematics courses.  By the end of it, you should be an expert at recognizing and graphing perpendicular lines.

Let's start with the perpendicular lines definition.  It specifically is talking about the relationship between two different lines who intersect at a 90 degree angle.  Two lines that cross at 90 degrees are said to be perpendicular to each other.  A good example of this is the familiar x-y axis.  The y-axis is perfectly up and down with absolutely no slant to the side, whereas the x-axis is perfectly left to right with absolutely no deviation to the vertical.  You can easily see that they form right angles where they meet.  However, it is important to realize that the lines themselves can go in any direction, not strictly up/down and left/right.  The only critical part is that they intersect at 90 degrees.

On a side note, a corollary to this definition is that the four angles created by the intersection of two perpendicular lines are all 90 degrees.  Comparatively, the intersection of two non-perpendicular lines results in the formation of two acute angles (less than 90 degrees) and two obtuse angles (more than 90).  This also demonstrates the geometric law which states that the sum of the angles around a point equals 360 degrees.  This is a useful rule to remember when solving geometric proofs.

So then, now that you understand what this word means and how to visually recognize it on a graph, you may then wish to prove that your two lines do indeed meet the criteria.  How would you even go about that?  How can you determine if two lines are perpendicular?  Well, to do this, you need to know the mathematical equations of the lines… or, more specifically, you need to know the slopes of the two lines.  (Recall that the slope of a line is most simply expressed as "rise over run", which represents the ratio of vertical change to horizontal change.  The slope of the line, often abbreviated by "m", is easily solved by comparing two ordered pairs, and then performing the slope calculation m = y₂-y₁ / x₂-x₁.)  True perpendicular slopes will have the following relationship:

m₁ = -1/m₂

In words, this means that the slope of the first line is equal to the negative inverse slope of the second line.  Looks a bit complicated, but it's not really.  Let's take a look at an example.

Consider the lines described by the equations y = 2x - 1, and y = (-1/2)x + 2.  Are these perpendicular?

This is an example of perpendicular lines.

If you want to get some practice graphing perpendicular lines, you can go ahead and plot these curves.  They are already expressed in standard form, so it is simple to determine the slopes and y-intercepts, and you can also readily generate a table of values to plot points along the lines.  That's a great way to show that you know how to graph the lines, and in the end, you would end up with two lines crossing at 90 degrees.  But that would take an awful lot of time on a test to find a solution that can be found much more quickly and easily.  Just consider the relationship that I explained, and see if it is true in this example.  You can see that the slope of the first line is m = 2, and the slope of the second line is m = -1/2.  This precisely fits the mathematical description of perpendicular lines.  You don't even technically need to graph it out to be able to answer this!  Of course, a wise plan of attack for solving this problem would be to check this relationship first, and THEN graph it out to show that you are correct.  This comes from personal experience.  Always check your work!  ;)

With this information, you should be able to see that all you need to know about lines are their slopes to be able to say whether they are perpendicular or parallel.  Recall, parallel lines have the exact same slope.  The trick when working with this type of question is to realize the the intercept values can be 2 and 3847234, or absolutely anything else at all.  They equation of the line may look completely and extraordinarily different for each, though the only important part of them is their m values.  Keep this in mind, and don't get intimidated by complex and scary-looking equations!

Another type of question might ask you to determine the equation of a line perpendicular to a given line through a specific point.  This takes a bit more work, but it is based on the same concepts.  Let's try a question like this.

Find the perpendicular line to the line y = 2x - 1 that goes through point (4,0).

Here's the approach I would take to solve this.

1. First, recognize that you are given one of the lines' equations, so from that you can easily find its slope. 
2. Second, from the first slope, you can use the perpendicular relationship to determine the slope of the second line.  
3. Third, since you now have the slope and a point that lies on the second line, you can substitute numbers into y = mx + b to solve for b, and then rewrite in in terms of x and y to give the final equation.  

I will leave the actual work for you to try yourself, but the line in this case is the same as above, y = -1/2x + 2.

A third type of question might ask you to determine the perpendicular bisector for a given line segment.  A bisector is a line that perfectly splits another line into two equal pieces, but it can slice through at any angle.  On the other hand, a perpendicular bisector is one the does this at precisely 90 degrees.  If you can first determine what the exact midpoint of your line segment is, you can then apply the approach that I outline above to solve this question as well.

There is one last important point that I would like to make about this topic, and it is about notation.  You are not incorrect to simply state that "line AB is perpendicular to line CD" (or whatever your lines are called), but the shorthand symbol to show this is an upside-down T shape, ⊥.  The keyboard character is called the "up tack", though this term is more applicable to lattice theory, type theory, and logic.  I believe it is more appropriate to simply call it the "perpendicular sign."  So, in this case, you would simply state your answer as AB⊥CD.  That's it.  It's much simpler!

Finally, I thought I would just throw in a bit of trivia that I came across while researching this topic.  Who knows… you might be able to impress your teacher!  The word "perpendicular" originally arose in the late 14th century, and its etymology shows that it came from the Latin word "perpendiculum", which means "plumb line", and "perpendicularis", which means "vertical, as a plumb line".  A plumb line was a simple device which was composed of a small weight on the end of a string, and when holding it up, gravity pulls the weight straight down and the string represents perfectly vertical.  In relation to the ideally perfectly horizontal ground, you can see how they came up with this term.  It's not overly useful information, but you never know where extra trivia might come in handy.

So, with that information, you should now know lots about this subject, and now have no problems graphing perpendicular lines or analyzing and identifying them in either graphs or equations.  There are several different variations to the questions that you may encounter, but if you understand the basics of what it is that defines two perpendicular lines, then you should have no problems in coming up with the appropriate solutions!  Please let me know in the comments below if you would like any further explanation or examples, and don't forget to +1 my post below and follow me on Twitter!  I'm @MathConcepts.  You can even click here to tweet about my post!  Be sure to visit my follow-up post that discusses a bit more of this concept of perpendicularity.

The Batman Equation

I can't verify that this is completely legit, but it certainly is very cool!  HardOCP has posted on their website a screen shot of an extremely elaborate mathematical formula, and its corresponding plot on a graph.  And, lo and behold, it's a Batman graph!

I wonder how long it will be before we have the Superman shield plotted out as well!  Though, I would imagine that as complicated as this Batman equation appears, it is probably just a piecemeal equation composed of lines and curves over specific domains.  The Superman equation would likely be far more complicated, as its shape is far more intricate.  If anyone has any free time and is able to put one together, send it my way and I'll be the first to post it for you!  :)

If ever there was a reason to learn how to make charts, this Batman graph is a pretty good one!  Though, judging by the complexity of the Batman equation (in the top part of the pic below), you would likely want to have a good graphing solver program.  Some of the most popular graphing calculators are made by Texas Instruments.  These TI calculators, such as TI 83, allow you to plug in your equations and have a big screen that will then display what your curve looks like.  These kinds of math graphing calculators would be invaluable if you need to work with complex mathematical formulas and their graphs.

*Edit: I'm surprised at the amount of traffic this blog posting generates!  When I first posted, I had no idea that so many people would end up searching for a Batman equation, and my site would rank so highly for it!  Check out this website as well for really cool visuals created by mathematical equations.  (Batman makes an appearance there as well.)  If nothing else, I hope that this demonstrates how useful math can be to do really cool things!  As I mentioned above, the equation shown above the Batman graph seems to be a piecemeal function, which utilizes a different equation over different domains of the function.  As you can see, piecemeal functions can produce some very interesting designs when graphed out.  Similarly, polar equations can produce very elegant designs as well!  A polar graph is based on an entirely different coordinate system, though it's just another example of how what seems to be just complicated strings of numbers and operations can come together to make something really interesting!  This general math concept should be taught to students early on, to hopefully catch their interest in pursuing further studies in mathematics!  After all, someone needs to develop the Superman equation!

Completing the Square

"Completing the square" is a method of expressing a quadratic function, and it is an especially useful form for graphing the expression.

If we have a function:

f(x) = (x+1)^2 + 3,

from the rules of graphing (1, 2, 3) that I have posted already, we can tell that this is a second degree polynomial (upright parabola), shifted to the left 1 unit, and up 3 units. However, if I were to ask you to graph:

f(x) = x^2 + 2x + 4,

that appears to be much more complicated. However, it describes the exact same curve, it is just written differently. "Completing the square" is the process used to convert this complicated and nasty form into the simpler form above.

To do this, you basically have to ignore the constant number (4, in this case) at first. So, to start, we look at x^2 + 2x. Now, we essentially do FOIL backwards to get it in the form (x+y)(x+y), or (x + y)^2... a little tricky, but with practice, you will get better at it and recognize certain patterns.

I will write as step by step below, and hopefully that demo will explain what I mean with all of this:

f(x) = x^2 + 2x + 4... re-writing to focus what we're doing:

f(x) = [x^2 + 2x] + 4... now reverse FOIL, determine the constant required in the square brackets to make it a perfect square. To keep things balanced, whatever number you add, you have to also take away:

f(x) = [x^2 + 2x + 2] + 4 - 2... simplify:
f(x) = (x+1)^2 +2

To get an x^2 and a 2x, we can see that the First term has to be just x, and the Outside and Inside terms have to be the same (because it's a perfect square) and add to 2x, which is 1x plus 1x, and which means that the y term in the general statement has to be 1. So, to get the x^2 + 2x correct, we have determined that is corresponds to (x+1)^2. Then, as noted, whatever number we have added to the expression to make it a perfect square, we also have to subtract from the number that we started with.

Hopefully this makes some sense to you. Let's try another example:

f(x) = x^2 + 10x + 37
f(x) = [x^2 + 10x] + 37
f(x) = [x^2 + 10x + 25] + 37 - 25
f(x) = (x+5)^2 + 12

Much easier to graph this than when it started. :) Here's another:

f(x) = x^2 + 14x + 20
f(x) = [x^2 + 14x] + 20
f(x) = [x^2 + 14x + 49] + 20 - 49
f(x) = (x+7)^2 -29

Let's try one more, a bit harder this time. Same logic applies.

f(x) = 4x^2 + 4x + 19
f(x) = [4x^2 + 4x] + 19
f(x) = [4x^2 + 4x + 1] + 19 -1
f(x) = (2x + 1)^2 + 18

Converting Point-Slope Form to Standard Form

I previously described how to obtain the equation of a line, and how to express that in both point-slope form and standard form. While both equations describe the exact same line, sometimes you may be asked to express the line in a specific way, and you need to be able to manipulate and rearrange the provided equation to make it look like the other form. I will show an example of how this can be done.  (Please hit the Like and/or Google +1 button at the bottom if you find this helpful!)

Reminders (refer to the posts linked above for more details)

Point-slope form looks like this:
(y-y1) = m(x-x1), which is the general way of saying y=mx+b

Standard form looks like this:
Ax + By = C

Example: Express the equation y=5x-10 in standard form. State the values for A, B, and C.

Basically, what you want to do is move all the x and y terms over to one side, and move the constants (terms with no variables) over to the other. Combine and simplify where possible. That's all there is to it. "A" will be the term left over in front of x, "B" will be with y, and C will be the value not attached to a variable.

y=5x-10

10=5x-y

So:
5x-y=10
A=5, B=(-1), C=10
(remember the standard form has a "+", so a "-" in your answer implies a coefficient of (-1).

Let's try another one:

Example:Express the equation y=(4/3)x+2 in standard form. State the values for A, B, and C.

This one works the same way, but there is something else that can be done, as I will demonstrate.

y=(4/3)x+2
(-2)=(4/3)x-y
So:
(4/3)x-y=(-2)
A=(4/3), B=(-1), C=(-2)
There is nothing wrong with this answer. It is properly rearranged, and the coefficients have been stated. However, usually it is a good idea to not have fractions (ie. have nothing in the denominator). So, to do this, you work our final answer a bit further, so that all the values are in the numerators.

(4/3)x-y=(-2)
Multiply all terms by 3, to remove it from the denominator of the first term. This gives:
4x-3y=(-6)
A=4, B=(-3), C=(-6)

Again, this answer describes the exact same line as the initial answer without the extra moves, so technically, they are both right. It is just a common convention to keep things in the numerator wherever possible.  Please hit the Like or Google +1 button below if this helped you.  :)

Converting from the Standard Form to the Point-slope form is basically just the reverse. Try it for yourself with these examples!

Manipulating graphs

In the next few days, I'm going to begin posting on graph manipulations. For any graphical expression, small changes to the expression can result in a very predictable change in the position of the graph. For instance, the graph can be shifted along the x-axis (horizontally), or along the y-axis (vertically). Similarly, the graph can be stretched or compressed along either axis as well. This may sound confusing, but the alterations to the equations are simple and easily recognizable. I will provide examples and go into detail in the coming days.

Graphing - Standard Form of the Equation

Just a short explanation for what is meant by "standard form" of the equation of the line. We have been looking at line equations in the form of y=mx+b. However, you may be asked to express this in standard form, or as a standard form equation.  Graphing standard form equations will give you the exact same line as graphing something expressed as y=mx+b... standard form is just a different way of displaying the equation.  (Please hit the Like button and/or the +1 button if this post is helpful for you!)

The general notation for a standard form equation is Ax + By = C, where A, B, and C are coefficients, and the x and y are the same variables we've been looking at but in a different position from what we recognize.

To express in standard form, you simply just rearrange the y = mx + b form such that you have x and y on the same side, equal to a number. Let's look at some examples:

Given that y = 3 x+ 5, standard form of this is 3x - y = (-5).

Given y = (1/2)x -15, standard form of this is (1/2)x - y = 15... also, if you don't want to have any fractions in your answer, you can multiply everything by the number in the denominator, such that we now get x - 2y = 30. Both expressions mean the same thing and will produce the same line. (In fact, convince yourself that no matter what you do to the equation, so long as you do it to both sides, the line is the same. eg. Multiply it all by 100, you get 100x-200y=30000... looks different, but it's not! Reduce it down and see for yourself!)

For graphing standard form equations, you still might want to go from standard form to the mx+b form, for which you may need to do a bit more math, but it's still quite straight forward.

Given 5x - 15y = 10, you just have to rearrange things to get y by itself on one side:
(-15y) = (-5x) + 10
y = (1/3)x - (2/3)...
and then you can see it is a line with slope 1/3 and y-intercept (-2/3).

Both types of equations mean the same thing. They are just expressed differently, and y=mx+b gives immediate information about the line without having to do a lot of work. However, you should be able to use both forms interchangeably.  Convince yourself that graphing standard form equations will give you the same line as graphing y=mx+b equations.  They just look different because the numbers are rearranged.  This should be obvious because if you start with a standard form equation, and convert it to y=mx+b and graph it, you have only rearranged things not added or removed anything.  You do not have a new line.

Also, from these equations, you should be able to tell that whenever you have an equation with 2 variables (x and y), and there aren't any exponents on either term, then you are dealing with a straight line. So while an equation in standard form may not immediately look like a straight line equation to you until it looks like y = mx + b, because it has an x and a y in it (without an exponent... exponents make the graph do cool things later), it is automatically a straight line.

(check this post for some additional pointers)

If you liked my post, you can help me spread the word by tweeting about it!  Also, please remember to click the Like button or +1 if it helped!

Graphing - Parallel and Perpendicular Lines

How can you determine if two lines are perpendicular?  How can you determine if two lines are parallel?  If you have two lines on a graph, and you have determined their equations or slopes, you may be asked if the two lines are parallel or perpendicular to each other.  These are two favorite questions of teachers and you will undoubtedly have to answer them!  Keep reading to find out how to easily answer them, and please remember to click the +1 button at the end.

Parallel lines are at the same angle and will never cross... like two railroad tracks. It doesn't matter what direction the lines travel. As long as they are going the same way, they are parallel. In mathematical terms, two lines are said to be parallel if they have the exact same slope.

Remember our equation for a line, y = mx + b.  Two parallel lines, each defined by their own equation, will have the same value for m, the slope.  So, y = 3x + 5 and y = 3x + 200 are parallel lines (they differ in their y-intercepts, but they have the same slope m).  You can plot this out for yourself quickly to see that this is indeed the case.

The opposite of parallel lines are perpendicular lines.  But how can you determine if two lines are perpendicular?  Perpendicular lines have a bit of a twist to them. Two lines are perpendicular if they cross (remember, any two straight lines that are NOT parallel will cross at only a single point.  They cannot ever intersect again unless they curve back on themselves, in which case they are not straight!) and they form a 90 degree angle, or rather, a T-shape. For example: The x-axis and y-axis are perpendicular to each other. Mathematically, if line 1 has a slope of m₁, then a perpendicular line 2 will have a slope m₂=(-1/m₁)... that is, it's slope will be the negative inverse of the first.

Try it out... y = 2x + 1 and y = (-1/2)x + 5... m₁ = 2 and m₂ = (-1/2). Check it out on the graph to see that they indeed form a 90 degree angle where they intersect.  If you don't believe that this is correct, go get a protractor and measure it for yourself!  You can then convince yourself that the relationship holds true.

Perpendicular graph

Also of interest to you: the symbol for "perpendicular" is ┴, an upside-down capital T, whereas the symbol for "parallel" is two vertical lines next to each other, like ||.  You would write AB┴CD to say that AB is perpendicular to CD, and similarly AB||CD to say that they are parallel.

So, I've shown you that to in order to determine if lines are parallel or perpendicular, all you need to know is their slopes!  If you know the equations of the lines, then this is only a matter of simply reading the m value, or at the worst, performing some basic mathematical rearrangements to find this info.  And once you have the numbers, you can easily see if either relationship applies.  This is much faster than manually creating a table of values and plotting out the graphs by hand!  Now, you should have no problem answering questions that ask you to quickly identify true parallel or perpendicular line relationships!  If this post helped solve your questions, please remember to do me a favor and click the +1 button below to help me get the word out!  Thanks!

Graphing - Equation of the Line

If you are given a line on a graph (or enough information to construct a line), you will likely also be asked to find the equation of the line. The equation of the line is unique to each line; that is, every line has a different equation. With it, you can readily tell the slope of the line, and you can calculate what the x-value is for any y-value on the line (and vice versa). It is very handy!

The most common and basic form of the equation of the line is:

y = mx + b

where m is the slope and b is the y-intercept (where the line crosses through the y-axis).

Another way to write it, which is more general is called the POINT-SLOPE FORMULA:

(y-y1) = m(x-x1)

where m is the slope, and x1 and y1 are the coordinates of a known point on the graph. (If you pay close attention, you can see that this way of writing it is really the same as the slope formula, but rearranged!) The version basically says 'give any point at all, and a slope, and you have enough information to draw the line."

Let's use the graph from the slope lesson to practice, using points (2,1) and (7,7). To get our final answer, we are going to have to do a couple of steps first. Let's use the y=mx+b equation. The steps we are going to do are:
1) Find the slope
2) Find the y-intercept
3) Write the equation of the line

Finding the slope is easy now, if you read the posting on slopes (I always leave numbers as fractions, instead of changing to decimals):

m = (y2-y1)/(x2-x1)
= (7-1)/(7-2)
= 6/5

The y-intercept is easy now... just plug numbers into y=mx+b, and solve for b. So, b=y-mx. We have our slope now, and for y and x, we just substitute in the coordinates of a single point (x and y MUST be from the same point!)

b=y-mx
= 1-(6/5) x 2
= 1-12/5
= (-7/5) (you have to do some fraction math!)

So now we can write the equation of our line!

y = mx+b
y = (6/5)x - (7/5).... (if we leave it like this, we can read the values for slope and y-int right from the equation!)
y = (6x - 7)/5

If we work through using the other formula... (y-y1) = m(x-x1)... we get the same answer, and we don't have to explicitly solve for b! First, solve for slope. Second, substitute in values for slope and x1,y1, and then rearrange!

slope = 6/5 (same thing we did above)

so (y-y1) = m(x-x1)..... (sub in slope, and point (2,1) for (x1,y1))
(y-1) = (6/5)(x-2)...
y=(6/5)(x-2) + 1...
y = (6/5)x - (6/5)2 + 1...
y=(6/5)x -12/5 + 1...
y=(6/5)x -(7/5)... (same slope and y-int)
y=(6x-7)/5

Same answer! That is the equation of the line. Now if you put in any value for x, you can say exactly what the value for y would be... if you wanted to know what y is when x is 1000, you can do that now! You will see that it will always be on the same straight line.

If we work through and do the same thing for the red line on the graph, with points (-6,2) and (-4,-5), we can get the equation for that line too! Let's try it this time using just the one formula.

(y-y1) = m(x-x1)
((-5)-2) = m((-4)-(-6))
(-7) = m(2)
m=(-7/2)

Now put that back into the same formula along with a point, and rearrange:

(y-y1) = m(x-x1)
(y-2) = (-7/2)(x-(-6))
y-2 = (-7/2)(x+6)
y= (-7/2)x + (-7/2)(6) + 2
y = (-7/2)x -21 + 2
y = (-7/2)x -19

(see slope (-7/2) and y-int (-19)... look at the graph, and you will see that makes sense! Negative slope, very low y-int!)

That's all there is to it! Just remember that there are a few steps to follow, depending on the formula you are using... basically, remember to always find the slope first, then the y-intercept, and plug directly into y=mx+b... or use the point-slope formula to find the slope using 2 points, and then resubstitute it back in with a single point and rearrange. It's not that complicated once you practice and understand what you are doing! Either method is going to give you the same answer, so pick your favorite and stick with it!

Graphing - Slopes

When you have a line on a graph, you will probably need to know its SLOPE. A slope on a graph means essentially the same thing as if you were talking about the slope of a driveway or the slope of a ski hill... it is a measure of how steep the driveway/hill/line is.

The minimum amount of information you need to find the slope of a line is the location of two points on the line. These could be endpoints for a line segment, or just points on a line that goes on forever. Since it's a single, straight line, ANY two different points that are on the line can be used and will give you the exact same slope as any other two points on the same line. Makes sense right? The slope is a property of the line, and all the points on the line make up the line.

The slope formula is easy to remember. The complicated way of saying it is 'the slope is the difference in height of two points on a line, divided by the difference in width of the same two points.' The much easier way is 'slope equals RISE OVER RUN.'

slope = rise / run

The rise (think rising = height) is the difference of the y-coordinates of two points on a line. The run (think about running on the street = horizontal) is the difference of the x-coordinates of the same two points.

Usually, slope is represented by the letter 'm.' So then, the slope equation can be written like:

m = (y2-y1) / (x2 - x1)

This appears to be a little more complex than the first one, but it means the same thing. The 1 and 2 are just names for the y's and x's. (They could be anything... A and B, or whatever.) The (y2-y1) means 'the difference between two y-coordinates, and the (x2-x1) means 'the difference between two x-coordinates.' IMPORTANT: Make sure that the point you use for y2 is the same point you use for x2, and likewise for y1 and x1. (Otherwise, you'll get the wrong slope.)

Another IMPORTANT thing to recognize: if the graph rises to the right, it is said to have a POSITIVE SLOPE. If it is falling towards the right, it has a NEGATIVE SLOPE. That means your slope value will have a positive or negative sign with its number. You can check that when you're done. (It's easy to make sign errors. It happens to everyone.)

This figure should hopefully clear things up.

Let's look at the black line first. Let's call the top point 'point 1' and the bottom point 'point 2'. So:

rise = (y2-y1) = (7-1) = 6
run = (x2-x1) = (7-2) = 5 **Notice that I didn't use (2-7)!

slope = rise/run = 6/5 or 1.2

That's it! Let's look at the red one now. It's a little trickier because of the negative signs, but you do the exact same thing. Point 2 on the left, point 1 on the right:

rise = (y2-y1) = (2-(-5)) = (2+5) = 7
run = (x2-x1) = ((-6)-(-4)) = ((-6)+4) = (-2)

slope = rise/run = 7/(-2) = (-7/2) or (-3.5)

Notice how the first graph had slope of (positive) 1.2 and was going up to the right, and the second one had slope (-3.5) and was falling to the right.

As long as you keep y2 and x2 coming from the same set of coordinates (and y1, x1), and you keep track of your signs, you'll get the right answer for your slope! And with the slope, you can do more complex things, like find an EQUATION that describes the line.