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Let's say we're running a basic algebra course and we're committed to showing proofs of everything we reasonably can. The development of equations of lines seems most straightforward in this sequence: (1) the standard (defining) formula $Ax + By = C$, (2) the definition of slope $m = \frac{y_2 - y_1}{x_2 - x_1}$, (3) the point-slope formula $y - y_1 = m(x - x_1)$, and then (4) the slope-intercept formula $y = mx + b$. That is: the definition in (2) after a few steps of manipulation leads to (3), which itself after a few steps of manipulation leads to (4).

However, if we consider presenting things in this sequence, there seems to be a dearth of interesting exercises we can assign at the point when we have formula (3) but not (4). We could just give a point and a slope and ask students to substitute them, but that is trivial/uninteresting. The half-dozen textbooks I've surveyed only ever use formula (3) as part of exercises (say, starting from two points) that ultimately lead to expressing the answer in the form of (4).

So: Are there any interesting (partially challenging, enlightening) exercises that make use of the point-slope form (3) without any reference or use of the slope-intercept form (4)? If not, then I suppose I should just present both formulas immediately together and then proceed to those exercises that convert from one to the other.

As a negative example, consider a sample exercise set from OpenStax College Algebra (Section 2.2). "For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form." (Various point-slopes and pairs of points are given.) That's not useful in our present case because we're at a moment when slope-intercept is not available. Likewise, the goal of this question is not to find applications or problem-solving uses (which are, somewhat embarrassingly, not part of the curriculum in question), but rather to directly exercise writing and manipulating the point-slope formula itself algebraically.

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    $\begingroup$ Does no one teach the two point interpolation form $y_{0}\frac{x - x_{1}}{x_{0} - x_{1}} + y_{1}\frac{x - x_{0}}{x_{1} - x_{0}}$? $\endgroup$
    – Dan Fox
    Commented Oct 22, 2017 at 9:53
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    $\begingroup$ @DanFox I expect it bit the dust with log/trig tables. The pressure has seemed to me to be to strip out all that is not used in calculus (and strip from calculus all that is not used in science). In fact, I seem to have students who were only taught the slope-intercept form, or at least that's all they know how to use. (And you only absolutely need one form of the equation of a line.) $\endgroup$
    – user1815
    Commented Oct 22, 2017 at 12:21

7 Answers 7

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The Fahrenheit and Celsius temperature scales are linearly related. A change of one degree Celsius is a change of 1.8 degrees Fahrenheit. The freezing point of water is 0° Celsius or 32° Fahrenheit. Find the formula for computing the Fahrenheit temperature $F$ given the Celsius temperature $C$.

[Probably most useful as an exercise in the US, where the students have heard of Fahrenheit.]

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Point-slope form is used all over the place in Calculus where the point and the slope (i.e. derivative) is exactly what you have in most situations. You can strip out any explicit calculus though and simply talk about rates. So something like the following:

A car is traveling 60 mph and passes mile-marker 13 at 3pm. Give an equation for the position $p$ of the car as a function of $t$ hours since noon. Answer: $p(t) = 13+60(t-3)$.

This is similar to ncr's answer where they use a rate of spending. But any rate can do. You can even do something like:

It's known that for each raise in a degree Fahrenheit of the average summer temperature, the population of Northern Pike in a lake decreases by 29. If at an average summer temperature of 83 degrees F, a fish count reveals 325 Northern Pike, give a model for the population $P$ as a function of temperature $T$. Answer: $P(T) = 325-29(T-83)$.

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I wonder why one needs to teach them as if they were separate formulas. When one knows one, the others are just rearranged versions.

Personally I find the definition of m easiest to memorize because you can just think of the triangle at the slope to derive the slope. Then you rearrange the formula for what quantity you need. This also helps for better understanding why the formulas related and why it is useful to be able to rearrange formulas.

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If you are teaching a class where a main theme of the class is translations of graphs, then you can also approach point-slope form this way:

A line with slope $m$ through the origin has equation $y = mx$.

Find the new equation for the line after it has been translated to the right by $x_1$ and up by $y_1$.

It used to pass through the origin. What point does it pass through now?

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One possibility: Given a point and fractional slope for a line, use the point-slope formula, and then write the line in standard form, and with no fractional coefficients. While not commonly seen in textbooks, this at least highlights the facts that: (a) any line described by slope and point can in fact be written in terms of the defining formula, (b) one should practice clearing fractions from equations, and (c) the attraction of standard form that any equation with rational coefficients can be expressed entirely with integers.

Example: Write the equation for the line through $(4, 8)$ and with slope $\frac{1}{2}$ in standard form with all integer coefficients. Solution: $y - 8 = \frac{1}{2}(x - 4) \Leftrightarrow 2y- 16 = x - 4 \Leftrightarrow x - 2y = -12$.

Now, several years after initially writing this answer, I discovered that this exercise is included in Martin-Gay, Intermediate Algebra, Section 3.5, which earns them extra points in my book.

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  • $\begingroup$ This is what I wound up doing in my college algebra course, and I'm pretty happy with it, so I'm selecting it as the accepted answer. $\endgroup$
    – Daniel R. Collins
    Commented Oct 28, 2020 at 4:27
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Maybe I'm not sure what you mean by "interesting" but is something like (from here):

Henry has \$125 in his pocket, but he is spending \$20 per hour at the arcade. If he leaves in 2.5 hours, how much money will he have left?

what you mean? It can be solved in a variety of ways but I think using the point slope form is most natural. There are other problems on the linked site that might be of interest, too.

For something a little more challenging, there are some AMC problems that might be interesting. For example Problem #11 on the 2010 AMC B Test, you'll find

A line with slope 3 intersects a line with slope 5 at the point $(10,15)$. What is the distance between the $x$-intercepts of these two lines?

A google search for "AMC problems point slope" might lead you to more such problems.

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    $\begingroup$ Unfortunately, I don't see that either of those problems require exercising the point-slope formula. The first is simply fixed arithmetic with no variable. The second can be done with a physical graph, or reasoning directly from the slope formula $m = \frac{\Delta y}{\Delta x}$ (ignoring the given x-coordinate, even). Comparing the x-intercept to the given point, for the 1st line we have $3 = \frac{15}{\Delta x} \Leftrightarrow \Delta x = 5$; for the 2nd line we have $5 = \frac{15}{\Delta x} \Leftrightarrow \Delta x = 3$; so the intercepts must be $2$ units apart. $\endgroup$
    – Daniel R. Collins
    Commented Oct 22, 2017 at 12:14
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I usually start with "A linear function is a function for which the increment of the function value is proportional to the increment of the argument", which is just a verbal version of (3) A typical question then would be:

For a linear $f$, we know that $$ f(2023)-f(2020)=1234. $$ Find $f(305)-f(5)$ in your head.

That seems to fit the bill because converting to (4) would involve working with one free parameter and some fractions while an intelligent use of (3) will give the answer immediately.

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