Slope: Free Response

Label the points and hold the order: $(x_1, y_1) = (-5, -2)$ and $(x_2, y_2) = (3, 4)$. Substitute, subtracting in the same order on the top and on the bottom.

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-2)}{3 - (-5)} = \frac{6}{8} = \frac{3}{4}.$

Both subtractions ran into a negative coordinate and both turned into an addition: $4 - (-2) = 6$ and $3 - (-5) = 8$. So the rise is $6$, the run is $8$, and once the fraction is reduced the line climbs $3$ units of height for every $4$ units across.

Read the slope as walking instructions: the denominator is the run and the numerator is the rise, so one step to the right moves $4$ across and $3$ up.

$(3, 4) \longrightarrow (3 + 4,\; 4 + 3) = (7, 7).$

Stepping the other way reverses both moves, $4$ to the left and $3$ down:

$(3, 4) \longrightarrow (3 - 4,\; 4 - 3) = (-1, 1).$

Both are on the line, and the second one can be checked against $P$: from $P(-5, -2)$ to $(-1, 1)$ the rise is $1 - (-2) = 3$ and the run is $-1 - (-5) = 4$, so

$m = \frac{3}{4},$

which is the slope the line already had.

Swapping the labels replaces $y_2 - y_1$ by $y_1 - y_2$ and $x_2 - x_1$ by $x_1 - x_2$. Each is the negative of what it was, so the new ratio is

$\frac{y_1 - y_2}{x_1 - x_2} = \frac{-(y_2 - y_1)}{-(x_2 - x_1)} = \frac{y_2 - y_1}{x_2 - x_1},$

since $\frac{-a}{-b} = \frac{a}{b}$: two minus signs, one above and one below, cancel each other. On the points of this question,

$\frac{-2 - 4}{-5 - 3} = \frac{-6}{-8} = \frac{3}{4},$

the same value as before. Nothing was risked by choosing an order, which is why the only rule is to keep the one you chose.

Mixing the orders is a different matter, because then only one of the two signs flips. Taking $y_2 - y_1$ on the top but $x_1 - x_2$ underneath gives

$\frac{4 - (-2)}{-5 - 3} = \frac{6}{-8} = -\frac{3}{4}.$

One sign changed and the other did not, so the ratio changed sign. That value belongs to a line falling to the right, the mirror image of the line you were given, so the error is not a small one: it reverses the direction the line reports.

The rise is the change in elevation and the run is the horizontal distance travelled while that change happens.

$m = \frac{\text{rise}}{\text{run}} = \frac{1030 - 820}{3500} = \frac{210}{3500}.$

Divide the top and the bottom by $70$:

$\frac{210}{3500} = \frac{3}{50}.$

A grade is the same slope written with one hundred underneath, so scale the fraction instead of recomputing anything:

$\frac{3}{50} = \frac{6}{100} = 6\%.$

Notice that the metres of rise and the metres of run cancel. A slope is a pure number, which is exactly why a grade can be posted on a sign with no unit attached to it.

A slope is the rise per unit of run, so the rise over a given run is the slope multiplied by that run.

$\text{rise} = -\frac{2}{25} \times 1500 = -\frac{3000}{25} = -120.$

The rise is negative because this section runs downhill, and $-120$ is a drop of $120$ m. The section starts where the first one finished, at $1030$ m, so

$1030 - 120 = 910.$

The road is $910$ m above sea level at the end of the second section. This is stepping along a line: one point was known, the slope supplied the rise for the run travelled, and the second point followed.

Compare sizes, not positions on the number line. Put both slopes over the same denominator:

$\frac{3}{50} = \frac{3}{50}, \qquad -\frac{2}{25} = -\frac{4}{50}.$

The first section changes elevation by $3$ units for every $50$ of run; the second changes it by $4$ units for every $50$. Ignoring direction, $4$ in $50$ is the bigger tilt, so the second section is the steeper one, even though $-\tfrac{4}{50}$ is the smaller of the two numbers. As grades, the comparison is $6\%$ against $8\%$.

That is the trap this part is built on. Ordering the two slopes as numbers puts the negative one below, and reading that as gentler is a mistake: the size measures steepness and the sign does not.

The signs carry the other half of the information. A positive slope means the road gains elevation as the driver moves forward along the surveyed direction, and a negative slope means it loses elevation. So a driver on the second section is descending, and descending the steeper of the two stretches.

Start at the left marked point, at $(-4, -2)$, and count in two moves.

The run comes first: from $x = -4$ across to $x = 4$ is $8$ units to the right. Now the rise: from $y = -2$ up to $y = 4$ is $6$ units up, and up counts as positive.

$m = \frac{\text{rise}}{\text{run}} = \frac{6}{8} = \frac{3}{4}.$

The formula agrees, as it must, since counting a triangle and subtracting coordinates are the same two differences:

$m = \frac{4 - (-2)}{4 - (-4)} = \frac{6}{8} = \frac{3}{4}.$

A smaller triangle on the same line works just as well. Between $(0, 1)$ and $(4, 4)$ the run is $4$ and the rise is $3$, and $\tfrac{3}{4}$ comes out again.

Start at the left marked point of $q$, at $(-3, 4)$, and count the run first: across to $x = 3$ is $6$ units to the right.

Now the rise. To get back onto the line you go from $y = 4$ down to $y = -4$, a drop of $8$, and a drop is a negative rise:

$m = \frac{\text{rise}}{\text{run}} = \frac{-8}{6} = -\frac{4}{3}.$

Counting the run to the right keeps the run positive, so the whole sign of the slope comes from the rise, and here the rise fell. That matches the picture: $q$ goes downhill from left to right.

Size and sign are separate readings. Comparing $\left|-\tfrac{4}{3}\right|$ with $\tfrac{3}{4}$, line $q$ is the steeper of the two lines, and it is the falling one.

Run each pair through the formula and watch which of the two differences collapses.

For $r$, the two heights are equal, so the rise is what collapses:

$m = \frac{6 - 6}{5 - (-2)} = \frac{0}{7} = 0.$

Zero divided by seven is a perfectly ordinary division with a perfectly ordinary answer. Line $r$ is horizontal, and $0$ is its slope.

For $s$, the two $x$-coordinates are equal, so it is the run that collapses:

$m = \frac{7 - (-1)}{4 - 4} = \frac{8}{0}.$

This is not a division with an unusual answer; it is not a division at all. No number multiplied by $0$ gives $8$, so nothing can be the value of this expression. Line $s$ is vertical, and its slope is undefined.

That is the difference in kind. One line has a slope, and the number happens to be $0$. The other has no slope for the formula to report. The two are opposite ends of the four cases, not neighbours, and the temptation to call a vertical line's slope infinite is worth resisting: infinity is not a number, and the division has no result rather than an enormous one.

There are three pairs, and each goes through the formula separately.

The first two points:

$m = \frac{2 - (-7)}{1 - (-2)} = \frac{9}{3} = 3.$

The last two points:

$m = \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3.$

The outer two, skipping the middle point, which doubles both the rise and the run:

$m = \frac{11 - (-7)}{4 - (-2)} = \frac{18}{6} = 3.$

The third pair is worth a second look. Its rise and run are both twice those of the first pair, and doubling the top and the bottom of a fraction leaves it alone, which is the whole phenomenon in miniature.

Take any two points of the line, $(x_1, y_1)$ and $(x_2, y_2)$, with $x_1 \neq x_2$. Nothing else is assumed about them, which is what makes the argument general.

Their run is $x_2 - x_1$, and since $x_1 \neq x_2$ that run is not zero. Their rise is $y_2 - y_1$. The constant-step property applies to this pair like any other, so

$y_2 - y_1 = k(x_2 - x_1).$

Substitute that into the slope formula. The run is not zero, so the division is legitimate and the common factor cancels:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{k(x_2 - x_1)}{x_2 - x_1} = k.$

The cancellation is the step where the points leave. Before it, the expression still mentions $x_1$ and $x_2$; after it, the whole expression is the single letter $k$, and $k$ was fixed by the line before anybody chose a point on it. So a second pair of points, sharing nothing with the first, runs through the same three lines and ends at the same $k$. That is what makes the slope of the line a phrase about the line rather than about the pair somebody happened to pick.

Notice what this is not. It is not the observation that swapping the two points you already hold leaves the value alone; that is a statement about one pair. Here the second pair may have no point in common with the first, and the conclusion still holds.

Take a vertical line, every point of which has the same $x$-coordinate $c$, and take two of its points, $(c, y_1)$ and $(c, y_2)$ with $y_1 \neq y_2$.

First, where part B stalls. The run is

$x_2 - x_1 = c - c = 0,$

and the proof divides by exactly that quantity. Dividing by zero is not a step that produces a wrong answer; it is not a step at all, so the argument simply does not reach a vertical line. The condition $x_1 \neq x_2$ was written into the statement for this reason.

Second, and more decisively, a vertical line never had the constant-step property to begin with. Suppose some number $k$ worked for it. Applied to the two points above, the property would say

$y_2 - y_1 = k(x_2 - x_1) = k \times 0 = 0,$

which forces $y_2 = y_1$. But the two points were chosen at different heights, and every vertical line has points at different heights. So no such $k$ exists, for any vertical line.

A vertical line therefore fails the hypothesis rather than the conclusion, and that is exactly what it means for it not to be a counterexample: part B never claimed anything about it. It also matches the picture. A vertical line has an undefined slope, so there is no single value for its pairs of points to agree on.

Contrast the horizontal line, which does meet the property, with $k = 0$: between any two of its points the rise is $0$, which is $0$ times the run. Its pairs all agree, on the perfectly good number $0$. Flat and upright are opposite cases here, not two versions of the same one.

Write the slope in terms of $t$ before anything else, taking $(2, -1)$ as point one and $(t, 5)$ as point two:

$m = \frac{5 - (-1)}{t - 2} = \frac{6}{t - 2}.$

Set that equal to the slope you want and solve:

$\frac{6}{t - 2} = \frac{3}{4} \quad\Longrightarrow\quad 3(t - 2) = 24 \quad\Longrightarrow\quad t - 2 = 8 \quad\Longrightarrow\quad t = 10.$

Check it against the two points. Through $(2, -1)$ and $(10, 5)$ the rise is $6$ and the run is $8$, and $\tfrac{6}{8} = \tfrac{3}{4}$.

Everything comes out of one expression:

$m = \frac{6}{t - 2}.$

The top is $6$ whatever $t$ is, because the two heights, $5$ and $-1$, are both fixed. So the sign of the slope is decided entirely by the sign of the bottom.

Positive. A positive divided by a positive is positive, so if $t > 2$ then $t - 2 > 0$ and $m > 0$. The converse holds too: if $m > 0$, then since the top is positive the bottom must be positive as well, so $t > 2$. Both directions hold, which is what lets this be stated as an exact condition rather than a one-way rule.

Negative. The same argument with the inequalities reversed gives $m < 0$ exactly when $t - 2 < 0$, that is $t < 2$.

Zero. A fraction is zero only when its top is zero. Here the top is $6$, so

$\frac{6}{t - 2} = 0$

would force $6 = 0$, which is false. No value of $t$ gives a slope of $0$. The geometry says the same thing: a slope of $0$ needs a horizontal line, and this line always joins a point at height $-1$ to a point at height $5$, so it can never lie flat.

Undefined. The one value the formula refuses is the one that makes the run zero, $t = 2$. Then both points sit at $x = 2$, the line is the vertical line $x = 2$, and its slope is undefined rather than very large.

So three of the four cases occur in this family and one, the flat case, is impossible for it.

Steepness is the size of the slope, and the size of

$m = \frac{6}{t - 2}$

is $6$ divided by the distance from $t$ to $2$, since the top never changes. Part B already separated the two cases: the run $t - 2$ is positive when $t > 2$ and negative when $t < 2$.

Steeper than any named number. Suppose somebody names a positive number $N$ and challenges you to beat it. Do not hunt for an example; build one from their $N$. Take

$t = 2 + \frac{6}{N + 1},$

which is one of the $t > 2$ cases. Then the run is $t - 2 = \tfrac{6}{N + 1}$, and the slope is

$m = \frac{6}{6 / (N + 1)} = N + 1,$

whose size is larger than $N$. Since $N$ was arbitrary, no line of the family is the steepest one: whichever you offer, moving $t$ a little closer to $2$ beats it.

Never flat. Flat means a slope of $0$, and the slope is $\tfrac{6}{t - 2}$, whose top is $6$ for every $t$. A fraction with a nonzero top is never $0$, so no line of the family is horizontal, no matter how $t$ is chosen.

What does happen is that the steepness shrinks. Pushing $t$ far from $2$ makes the distance from $t$ to $2$ large and the size of the slope small: at $t = 602$ that distance is $600$, and

$m = \frac{6}{600} = \frac{1}{100}.$

So the lines get as close to flat as you please without ever arriving, and arriving is a different thing from getting close. The same is nearly true at the other end: no line of the family that has a slope is vertical, however steep it gets, and the single value $t = 2$ that does give a vertical line is exactly the one where the slope stops existing.