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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). Of course, 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.

Edit: Compare to 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.) Of course, here we consider the case when slope-intercept is not available. The point of this question is not to find applications or problem-solving uses (which is, 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 Oct 22 '17 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.) $\endgroup$ – user1527 Oct 22 '17 at 12:21
  • $\begingroup$ @DanFox: I have never seen that before. $\endgroup$ – Daniel R. Collins Oct 22 '17 at 12:21
  • $\begingroup$ I suggested a tag edit to add undergrad education and secondary education; I guess those are the appropriate terms. $\endgroup$ – Tommi Oct 23 '17 at 7:33
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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 Celcius temperature $C$.

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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 seen in any textbook, 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: $y - 8 = \frac{1}{2}(x - 4) \Leftrightarrow 2y- 16 = x - 4 \Leftrightarrow x - 2y = -12$.

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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 Oct 22 '17 at 12:14
  • $\begingroup$ @DanielR.Collins My understanding was that you wanted problems that could naturally be solved with your formula (3) and would be less naturally solved with (4). As Michael E2 points out, there is almost no difference between formulas (2) and (3) and so almost any problem that can be solved with (3) can also be solved with (2). $\endgroup$ – ncr Oct 22 '17 at 15:47
  • $\begingroup$ @ncr See the recent edit to the question; I hope that is clarifying. $\endgroup$ – Daniel R. Collins Oct 22 '17 at 19:20

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