# Should I be teaching point-slope formula to high school algebra students?

I'm student teaching this semester, and so far I'm loving it! Our next section in the book teaches point-slope formula, and my cooperating teacher (a 24-year veteran teacher) is convinced that point-slope formula serves no purpose. He said he has been at a constant "battle" with the other teachers at the high school because there is no purpose. At first, I challenged him, and asked, "what if you have a point far off from the origin?".

Well, the question would provide you with a point and a slope. He said you can just plug those into slope-intercept form (which he loves) and solve for b (the y-intercept). I did some googling and found that the point-slope form reappears in Calc 2 and Calc 3 with polynomials and linear approximation functions. When I showed that to him, he said, "well by the time they get there, they'll just learn it new". So he's STILL not convinced that we should be teaching point-slope formula to these students!

I even challenged him by asking, "isn't it important for these students to see multiple ways to write equations for lines?". No, slope-intercept is all they need to get through the class. "Well, if you're given a point and a slope (say x1,y1 and m), you could plug them into the equation to get y-y1=m(x-x1) and type that straight into desmos and get a line without having to do any work". Well, my technology argument was a no-go as well. So I'm at a loss.

EDIT: After reading a few replies, I showed him how simple it is to derive the point-slope formula from the slope form m=(y2-y1)/(x2-x1), and then from there you could derive the slope-intercept form, but his response was, "or they could just know slope-intercept form and use it every time".

Am I crazy in thinking that point-slope formula is important for these students in 9th grade honors algebra? How can I convince my cooperating teacher?

• I teach college, not high school, so my expectations might not be appropriate here -- but I see your mentor's attitude as being way, way off base. The slope-intercept form and the point-slope form are trivial variations on one another. A good student should be able to independently derive one from the other, without explicit instruction. Although a weaker student might not be able to figure out the correspondence from scratch, it would be totally unacceptable IMO for such a student to emerge from even a basic Algebra 1 class without being able to handle both forms. – Ben Crowell Nov 2 '15 at 23:35
• How can I convince my cooperating teacher? He's been teaching for 24 years. Obviously you're not going to change his mind. Ignore him and do the right thing yourself when you have your own classroom. – Ben Crowell Nov 2 '15 at 23:36
• Students should not only know point slope form, but should be able to write a proof showing they are equivalent. 24 years of repetition means he's very experienced at doing things wrong. Perhaps you can have the calculus teacher in your school send him an e-mail asking him to make sure the kids learn point-slope form so they are prepared for calculus. – Greg Nov 2 '15 at 23:41
• Thanks for the words Ben. This semester has been frustrating. I think you're right though. I've never won any arguments (though I admit, I'm young and I'm sure a lot of my opinions are subject to change with experience), but it's probably not worth trying to convince him otherwise. A lot of the things we've taught have been inaccurate for these students. For example, teaching half-life as an example of inverse variation. For those who don't remember, half-life is exponential decay whereas inverse variation is of the form y=k/x where k is some constant. – Wmol Nov 3 '15 at 4:44
• For whatever it's worth, the point-slope formula can be milked in precalculus classes by showing how the precalculus notions of horizontal and vertical shifts lead one from the middle school idea of direct variation to the point-slope formula. Start with $y=mx,$ and replace $x$ with $x-a$ (shifts right by $a$ units) and replace $y$ with $y-b$ (shifts down by $b$ units). – Dave L Renfro Nov 3 '15 at 21:00

As someone who teaches calculus to college students, I expect my students to have seen point-slope form. We just start using it (because it's the right way to talk about tangent lines and linearization) without teaching it, because we consider it part of the standard algebra curriculum, so students who haven't seen it are at a disadvantage.

Further, students who haven't seen the idea that there are multiple ways to write equations for lines are at a bigger disadvantage, because they have to grasp the idea that there are multiple representations, and a particular new representation, and the actual new course material, all at once.

Point slope form emphasizes the actual meaning of slope.

Literally,

$$y - b = m(x -a)$$

Says

"The change in the outputs ($y-b$) is equal to the slope ($m$) times the change in the inputs ($x-a$)".

Translating between a verbal statement like this and an equation is essential. Understanding slope is essential. Point slope form of a line is essential.

It links to transformations of functions, which should be the main story in precalculus.

It is, ridiculously, the first order taylor expansion of the function about $x=a$.

More than any technical details about this particular problem, however, this focus on one form betrays a whole way of thinking. Namely, that we shouldn't confuse students, and we should give them dependable algorithms that always work. From this perspective, your collogue may be correct, and only using one form may be "easier for the students". But if the goal is to get the students to think, understand, explain, explore, then they are very incorrect.

• Hm interesting. Maybe thinking about it this way would help students create a conceptual understanding between tables and graphs. Maybe this is because I'm exhausted, but how is the change in outputs written as (y-b)? Similarly, how is the change inputs (x-a)? Rather than something like x2-x1, or x sub n - x sub n-1 (hope this makes sense, I'm not sure how to format this). I assume b is the y-intercept and a is the x-intercept? Thank you! – Wmol Nov 4 '15 at 5:26
• @Wmol Sorry, $(a,b)$ was the "point" and $m$ was the "slope" in this example. – Steven Gubkin Nov 4 '15 at 17:18
• I'm a little confused by your explanation in quotes. Do you mean $\Delta y = m \Delta x$. If so I find the slope intercept more clearly shows this with $y=mx$ translated up or down by $c$ (with $c$ meaning both the y-intercept and the constant of integration). I prefer to think of the 'point-slope' form as simply the standard form translated by (a,b). – Richard Nov 6 '15 at 12:30
• I know the slope is $m$, I know $(a,b)$ is on the curve. I want to take an arbitrary point $(x,y)$ on the curve, and figure out a relationship between $x$ and $y$. The only thing I can do is use the two facts I know. When the input changes from $a$ to $x$, the output changes from $b$ to $y$. Since slope is defined as the number by which you multiply changes in inputs to get changes in outputs, we have $y-b = m(x-a)$. – Steven Gubkin Nov 6 '15 at 17:13

As someone else teaching calculus and higher math to college students, I use point slope form repeatedly:

• In the/a definition of derivative, I use point-slope form. I have students think about $$y_1-y_2=m(x_1-x_2)$$ rewritten as $$f(x_1)-f(x_2)=m(x_1-x_2)$$ rewritten as $$f(x+\Delta x)-f(x)=m(x+\Delta x-x)$$ rewritten with limits to describe the slope $$m$$ as the derivative.
• In linear approximation, I likewise use point-slope form.
• I make my students apply these ideas using specific functions, like $$f(x) = x^p$$. Yes, it's just applying the definition of derivative but I make them draw pictures to see that it's the point-slope form of a line coming up again.
• In discussing the fundamental theorem of calculus, I use point-slope form.

I tell my first-semester calculus students that if they have point-slope form down cold they can probably pass the whole class. I also tell them that it's the beautiful and lazy way to write down the equation of a line, and I don't want to see slope-intercept for the first five weeks. If they still love slope-intercept by the end of five weeks they can go back... but most don't :)

• That's the first thing that popped into my head is an entry point to the derivative... – corsiKa Nov 2 '15 at 21:54
• Awesome, thank you so much. So is it a limit as Dx approaches zero? It's been awhile since I've taken calc. – Wmol Nov 4 '15 at 5:17

I am someone whose mind seems to work similar to your cooperating teacher. Why learn two forms of something when you can learn one and manipulate it? I made it through all of algebra 2 never learning the quadratic formula. Instead, I learned how to derive it by completing the square, and I started every test by re-deriving it!

I think its essential that a student who learns point slope learns that it's nothing magical. It's just solving for a variable, as you noticed. However it has two major features:

• There's less algebra, more arithmetic.
• Its in a form which mirrors the way a problem often reveals itself.

From a teaching perspective, the former might be a disadvantage. This might be what your cooperating teacher is picking up on. The more you practice algebra, the better you are at it.

However, now that I am no longer a student, I find problems often present themselves in the form of a point and a slope. Knowing that there is a quick arithmetic way to solve these sorts of equations draws my attention to them more.

It's also very helpful as a computer programmer. My software can't always solve an equation for a variable, but it can do arithmetic. I rely heavily on the point-slope formula for these situations.

Personally, I went many years not teaching the point-slope formula (college algebra and remedial algebra), and I just flipped on the issue this past weekend. Here's why: In making sure that I could justify all the basic graphing principles, I found that this formula was the easiest way to establish that lines can indeed be written as linear equations (restricting to defined slope for simplicity):

Theorem: Assume line L has defined slope $m$ and some point $(x_1, y_1)$. There is a linear equation such that any point on the line is a solution to the equation, specifically: $y - y_1 = m(x - x_1)$. Proof: Given any other point $(x_2, y_2)$, we know $m = \frac{y_2 - y_1}{x_2 - x_1}$ because slope is constant on a line. Check that the points are solutions to this equation: For the first point we get $y_1 - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x_1 - x_1) \Rightarrow 0 = 0$, while for the second point we get $y_2 - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x_2 - x_1) \Rightarrow y_2 - y_1 = y_2 - y_1$.

Likewise, I discovered that the point-slope formula was about the easiest way to get started on several of the other proofs, working through the basic graphing principles from an axiomatic basis. It was just yesterday that I rewrote all my lecture notes to include it going forward -- and this quick proof that lines are indeed linear equations, whereas for the prior decade I was just hand-waving the issue intuitively.

• This is interesting. Would it be possible to see your lecture notes? If it's inconvenient then never mind, but I'm trying to wrap my brain around proving these ideas to algebra students. I know the age gap is pretty different (College vs 7th and 8th graders in adv. 9th grade math), but the book we're using is a remedial college textbook, so I think the students could probably handle it. – Wmol Nov 4 '15 at 5:22
• @Wmol -- Email me at dcollins at superdan dot net. – Daniel R. Collins Nov 4 '15 at 5:29
• I also, personally, remember point-slope using $m = \Delta y/ \Delta x$ (the classic classroom mnemonic of "rise over run"). Since I know a point and slope are enough to determine a line uniquely, I pick a point $(a, b)$ on the line and rewrite the above equation as $m = (y-b)/(x-a)$. – Benjamin Dickman Mar 24 '18 at 4:24
• @BenjaminDickman: Right, how I recall it (and recommend to my students) is as the fractionless $\Delta y = m \Delta x$. – Daniel R. Collins Mar 24 '18 at 13:00

I remember being slightly giddy the first time I was told that the equation of the line with x-intercept a and y-intercept b is $\frac xa + \frac yb = 1.$ I love this equation and the generalizations it implies. My guess is that less than one percent of the algebra students in a school would be impressed with this equation. So, occassionally, I teach to the one percent. Not really for them, but because I get a kick when that one student gets that look on his/her face.

I got to thinking about this. If you can find two "nice" points $(x_1, y_1)$ and $(x_2, y_2)$, then the equation of the line passing through those two points is $$\frac{x-x_1}{x_2-x_1} + \frac{y-y_2}{y_1-y_2} = 1$$

• Your first equation really does have a beautiful symmetry to it. The wonderful thing about linear equations is that all you have to do is check that it has 2 points correct ((a,0) and (0,b)), and you can be guaranteed that the rest works. Interesting visualising a line as y=-x+1 scaled appropriately rather than y=x translated. Not so good with a (0,0) intercept though. As for your second equation, though I can see that works (checking at both points), it does not seem as intuitively true. – Richard Nov 6 '15 at 11:59
• Just read Daniel R. Collins comment on another answer, and can see how useful this form also is for moving to linear algebra: Ax + By = C implies a=C/A, b=C/B. Also starting with this form allows a smoother transition to equations of ellipses. What do you think of $\frac{x^2}{a} + \frac{y}{b} = 1$ as a general form of a parabola? I might need to experiment with that. – Richard Nov 6 '15 at 12:23
• @Richard : Like I said, I love the generalizations that it implies. I'm not really that fond of the second of my equations. I got it by translating the origin of the first to $(x_1, y_2)$, the vertex of the right triangle whose hypotenuse is between $(x_1,y_1)$ and $(x_2, y_2)$. That's what I found to be interesting. – Steven Gregory Nov 6 '15 at 14:02

It can't be that important because I have never heard of it before (and teach at university level). I believe textbooks will have some things because they provide an easy form of question rather than because they actually matter.

Edit: The more I think about this, the more I agree with your teacher, in that I think it is unhelpful to teach both given that only one is needed. What I mean is, the 'point-slope' equation is the same equation as the 'slope-intercept' equation. As such, I think it is unhelpful to teach one concept as if it is two different ones. I believe that telling students incorrectly that these are different will encourage their natural tendency to think of the equation of a line as a string of characters, rather than a mathematical statement that is true for points on the line and false otherwise.

Also, emphasising the difference makes out that there is an important difference, when all the difference is is basic algebra. I would think this sends the message that the algebra is difficult and should be expected to be difficult (already far too wide-spread an opinion). Instead we should be teaching that the algebra is not difficult and is not the important feature.

My suggestion, then, would be to teach both ways of writing the equation, showing that one or other may be more convenient depending on the information you have available, and can help in different ways to aid intuition. Emphasise that the two only differ by rearranging the formula, and that they can easily move between the two to get the viewpoint they find helpful. Also make it clear that mathematical convention is to write $y=mx+c$, and so they should give their answer in this form, whichever way they arrive at it. (Before I get objections to this last point: if a student wrote something else I would consider it laziness, and if I saw anything else in a paper without reason I would be very surprised.)

• The title question was 'Why should I be teaching this?' and the main question now asks 'Am I right to think this is important?' My answer is 'no, it's not important, and you can teach it but there isn't that much of a good reason'. Looking at the wikipedia page, I'd say some people have really got over-zealous and missed the point that the equations are all the same. – Jessica B Nov 4 '15 at 8:50
• @BenjaminDickman As I've now expanded, it is because students think the same equation is different that I say you should not teach it as different. They need to practice the skill of seeing why the two are the same, not the skill of following a procedure that magically turns one thing into another. – Jessica B Nov 4 '15 at 9:37
• mathematical convention is not just to write y=mx + c across the board. It depends on what you are doing. If you want the formula for the function for that line, that may be so, but if you want to define slope (y - y0) = m (x - x0) is a better convention to start from. And if you get two points on a line and need to figure out the equation for it, it is much easier to start with this form. I grew up on y=mx+b and have always preferred it, but this other form of the same equation is as much a convention. But you are right that students should learn how either is easily derived from the other. – AgapwIesu Nov 4 '15 at 17:48
• Also, an example of another form: In the context of solving systems of linear equations, then the convention is to write in the general form Ax + By = C. This is included in our remedial/elementary algebra classes, so the emphasis there is actually on graphing directly from this form via intercepts. This then supports solving by graphing and writing matrices later on. – Daniel R. Collins Nov 5 '15 at 23:35
• @DanielR.Collins : I like to point out that the equation of a line parallel to the line Ax + By = C is Ax+By=C' and the equation of a line perpendicular to the line Ax+By=C is Bx-Ay = C'. – Steven Gregory Jun 19 '17 at 19:31

Point slope form teaches a very important concept in a very simple way - that of translation. When you get into quadratics or any other equation, you can move the equation around by taking a standard equation and bumping the origin around. Point slope is the simplest way to do this. It takes a line of a particular slope which passes through the origin and translates it on the graph by moving the origin point to (a, b).

This technique (replacing $y$ with $(y - b)$ and replacing $x$ with $(x - a)$ to get a translation of the graph) is useful to every other form of graph translation and applies to every other equation.

Here is a parabola shifted over so the vertex is at $(2, 3)$:
$$(y - 3) = (x - 2)^2$$

Here is a circle of radius $5$ whose center is at $(-2, 7)$:

$$(x - -2)^2 + (y - 7)^2 = 5^2$$

I would also argue that the most important thing you can learn in a math class is how to connect different ideas, and see the same ideas in different ways. Solving problems is an important but secondary consideration in a math class.

This is a follow-up and add-on to johnnyb's excellent and important answer.

One of the most important themes in Algebra 2 and higher-level classes is that of function transformation. This theme runs throughout virtually every topic in those courses. Consider the following function types:

• Linear functions: $y = A(x-x_0) + y_0$
• Quadratic functions: $y = A(x-x_0)^2 + y_0$
• Exponential functions: $y = Ab^{x-x_0} + y_0$
• Sinusoidal functions: $y = A\sin(x-x_0) + y_0$
• Logarithmic functions: $y = A\log_b (x-x_0) + y_0$

All of these functions have the same basic form: take a "toolkit" function ($y=x$, $y=x^2$, $y=b^x$, $y=\sin x$, or $y=\log_b x$) and perform (in sequence) the following transformations:

1. shift it horizontally
2. rescale it vertically
3. shift it vertically

Now you might have noticed that "rescale it horizontally" is missing from the list above. For linear and quadratic functions, there's a good reason for that: if $f(x)=x$ then $f(k\cdot x) = k\cdot f(x)$, which shows that a horizontal "compression" by a factor of $k$ is indistinguishable from a vertical "stretch" by the same factor; similarly if $g(x)=x^2$ then $g(k\cdot x) = k^2\cdot g(x)$, so a horizontal "compression" by a factor of $k$ is the same as a vertical "stretch" by a factor of $k^2$. (Incidentally, this is one of the reasons why when students are learning about quadratic functions in vertex form they often mis-recognize "make it taller" as "make it skinnier".) Even for exponential functions, a horizontal rescaling can be absorbed into a change of the base, so you don't (strictly speaking) need them. It's really only for trigonometric functions that the horizontal scale factor plays a role that can't be accounted for by one of the other parameters.

In high school, transformations are usually taught as a discrete topic, confined to an early chapter of Algebra 2 and only occasionally mentioned again, but I really think it is one of the most important throughlines of mathematics at the secondary and early postsecondary level.

In teaching university-level Calculus, for example, I will often begin the semester with a set of questions like this:

1. Find a formula for a linear function that passes through $(2,5)$ and $(9,15)$.
2. Find a formula for an exponential function that passes through $(2,5)$ and $(9,15)$.
3. Find a formula for a quadratic function that has its vertex at $(2,5)$ and passes through $(9,15)$.
4. Find a formula for a trigonometric function that has a minimum at $(2,5)$ and whose first maximum after that minimum is at $(9,15)$.

These are meant to be review problems, although students always struggle with questions 2-4. Needless to say, every student can answer question #1. Most will use point-slope form; a few will use slope-intercept form, and then rewrite their answer in point-slope form.

However, when we move on to #2, most students go through elaborate gyrations to find the values of $A$ and $b$ so that $y=A\cdot b^x$ passes through the given points. Then I show them that the function can be described as

When $x=2$ set $y=5$, and multiply $y$ by $3$ every time $x$ increases by $7$

and that this translates directly into the formula $$y=5\cdot 3^{(x-2)/7}$$

Students are usually astonished that it can be done that easily, at which point I remark that this is really the difference between slope-intercept form (which requires you to describe the line in terms of its $y$-intercept, regardless of whether that point is actually important for you) and point-slope form (which allows you to directly use the points you are given). We then spend the next couple of days approaching the rest of the problems using similar methods. This not only works well as a review of prerequisite material but also sets the stage for finding the equations of tangent lines, and (much later) Taylor series.

The point-slope form is (or has the potential to be) students' first encounter with transformational thinking and lays the groundwork for these more advanced topics. I think it is essential material for high school algebra.

I have always argued for teaching it.

1) Like others have mentioned it has a concrete connection to the slope formula. This can be a nice teaching moment on deriving formulas.

2) It relates to the standard form of many other types of curves. Students who have a good grasp of the different forms of lines tend to do better when we introduce the conic sections, and I think that they're recognizing some of those similarities.

3) It is a good stepping off point for talking about transformations. It's similar to other common forms because those forms show the transformation form the origin for that family of curves.

4) It is helpful for some students who struggle with abstract concepts. When teaching this in high school, it is helpful for some students to be encouraged to "get something on paper." This form requires the least specific information to get some kind of equation down on paper.

Tangent at a point. The (currently) most-voted answer mentions this, but not in the multivariable context.

The easy generalization of $$L(x)=m(x-a)+K$$ to the multivariate context is $$L(x,y)=m_x(x-a)+m_y(y-b)+K$$ (and so forth for more variables, eventually you get into matrix/vector version...)

Naturally, you could also solve some $ax+by+c=z$ here, but it gets less and less connected to reality.

However, I will grant that for most high-school teachers in the US this will not seem the most compelling argument, since the small percentage of students who go on to this will probably survive no matter what version you use in 9th grade.

I see many good points here, addressing issues going forward. But I have a lower-level reason. I teach point slope form in preference to y=mx+b because it has better flow.

When they use y=mx+b, I have seen students plug in a point to get b, and then stop. They forget to give the equation of the line, which is what they were trying to find. When they use point slope form, it flows better.

I try to help students see why they would want to use each form, and how one form can transform into another.

As a student, I was taught all three major line formulas: point-slope, slope-intercept, and two-point (and even exposed to the intercept, intercept oddity). I always devolved to just using point slope as it feels smaller and more intuitive. Still to this day is how I think of lines. I think there is a physical intuition to it as well: initial condition and rate of change. But I was forced to be exposed to all of the equations. And I agree that, provided you remember them, the other equations work faster for problems suited to them.

As for the teacher, of course, you should let him do things how he wants when you are the assistant and the class belongs to him. When you get your own dojo, you can teach karate how you like it.