The Coordinate Plane: Free Response

An ordered pair records two journeys from the origin, and the convention fixes which one is written first: across the page, then up or down. So the first slot always takes the left-or-right move and the second always takes the up-or-down move.

The sign of each number is what records the direction of its own move. Right and up are the positive directions, left and down the negative ones, exactly as on the two number lines the axes are built from. Reading the three descriptions with that convention:

$M = (8, -6), \qquad N = (-6, 8), \qquad R = (0, -9).$

Point $R$ is the one worth pausing on. It is described with no sideways movement, and a move of nothing across is recorded as $0$ in the first slot. The zero is not optional: the pair must still carry two numbers, because a point in the plane needs both a sideways position and a height before it is pinned down.

A quadrant is settled by two facts and nothing else: which side of the y-axis the point is on, and which side of the x-axis. Those two facts are exactly what the two signs record.

Point $M$ has a positive first coordinate, so it is right of the y-axis, and a negative second coordinate, so it is below the x-axis: that is the lower right. Point $N$ has the opposite pair of signs, so it is left of the y-axis and above the x-axis, the upper left:

$M: (+, -) \Rightarrow \text{Quadrant IV}, \qquad N: (-, +) \Rightarrow \text{Quadrant II}.$

Point $R$ never moves sideways, so it stays on the vertical line through the origin, which is the y-axis. This is the place the naming trips people up: the coordinate that is zero is the x-coordinate, and yet the point sits on the y-axis, because a zero across leaves the point on the up-and-down line.

And a point on an axis is in no quadrant. The quadrants are the four regions the axes cut the plane into, strictly between them, so a point standing on a boundary belongs to neither region it separates.

The two points in the stem are already a swapped pair, so test the claim on them first. Point $M$ is $8$ across and $6$ down while point $N$ is $6$ across the other way and $8$ up, and those are plainly different places on the page:

$M = (8, -6), \qquad N = (-6, 8).$

So for these two the claim holds up. That settles nothing in general, because a claim with the word always in it is a claim about every point at once, and one point that refuses it is enough to bring it down.

So ask what the swap actually does. It sends a general point $(a, b)$ to $(b, a)$. Two points are the same point exactly when they agree in the first coordinate and agree in the second, so $(a, b)$ and $(b, a)$ are the same point exactly when $a = b$: the first coordinates agree only if $a = b$, and if $a = b$ then both coordinates agree and the swap has done nothing at all.

That is a genuine family of points, not a curiosity. The point $(-7, -7)$ is unmoved by the swap, and so is the origin, since $0$ and $0$ are equal. Every point whose two coordinates differ does move, which covers almost everything, and that is presumably what the student had in mind. The claim as stated is still false, because always admits no exceptions.

Check the alignment before reaching for arithmetic. Both points carry the same second coordinate, $-4$, so they stand at the same height and the segment joining them runs straight across the grid. Along that segment nothing but the sideways position changes, so the pair behaves exactly like two marks on a single number line.

On a number line the gap between two positions is the absolute value of their difference, so subtract the coordinates that differ, the x-coordinates, and take the absolute value:

$d = |8 - (-9)| = |8 + 9| = 17.$

The inner minus signs are doing two different jobs and are easy to run together: one is the subtraction, the other belongs to the coordinate $-9$. Subtracting a negative adds its opposite, which is why the two numbers combine rather than cancel.

Counting the gridlines checks it. From $x = -9$ to the y-axis is $9$ units, and from the y-axis to $x = 8$ is another $8$, so the two points are $17$ units apart, and the answer is a positive length in grid units.

Sort the three pairs by what they have in common, since that is what decides whether a subtraction measures anything.

$P$ and $Q$ share the height $y = -4$, which part A already used, and they are $17$ units apart. $P$ and $T$ share the column $x = -9$, so the segment between them is vertical and the y-coordinates are the ones that differ:

$d = |9 - (-4)| = 13.$

$Q$ and $T$ are the odd pair out. Their x-coordinates differ, $8$ against $-9$, and so do their y-coordinates, $-4$ against $9$, so the segment joining them is slanted and lies along no gridline. Neither subtraction measures it. Subtracting the x-coordinates gives $17$ and subtracting the y-coordinates gives $13$, but those are the two stretches of a staircase route from one point to the other, not the straight-line gap, and going across and then up is a longer journey than going straight.

So the rule has a condition attached, and the condition is the whole of it: subtracting coordinates gives a distance exactly when the two points agree in the other coordinate. Where they agree in neither, the distance is a real length that nothing in this lesson measures.

Look at what a plain subtraction actually returns. Taking the x-coordinates of $P$ and $Q$ in the two possible orders gives two results that differ only in sign:

$8 - (-9) = 17, \qquad -9 - 8 = -17.$

Both are reporting the same $17$ units of separation. What the sign adds is direction: it says which of the two points lies further right, and it flips when you list them the other way round.

A distance is not asking that question. It asks how far apart, which is a count of units, and a count is never negative. So the direction is information the answer does not want, and the absolute value bars are what remove it, since $|17|$ and $|-17|$ are both $17$.

That also disposes of the ordering. Since the two orders differ only by a sign, and the bars discard exactly that sign, both orders arrive at the same length. This matters more than it looks: neither of two points is the first one in any real sense, so a measurement that depended on which one you happened to write down first would not be measuring the pair at all.

A side length is a gap between two coordinates here, not something to be measured with a tape, so start by picking two corners that share a coordinate.

The bottom side joins $(-10, -6)$ to $(2, -6)$. Both stand at the height $y = -6$, so that side is horizontal and its length is the gap in the x-coordinates:

$l = |2 - (-10)| = 12.$

The left side joins $(-10, -6)$ to $(-10, 2)$. Both stand in the column $x = -10$, so that side is vertical and its length is the gap in the y-coordinates:

$w = |2 - (-6)| = 8.$

Both subtractions had to cross zero, because the marker post stands inside the plot, and that is where a sign is usually dropped. The counting check keeps it honest: the plot reaches $10$ m one side of the post and $2$ m the other, which is $12$ m across, and $6$ m below the post plus $2$ m above, which is $8$ m up.

The fence runs along all four sides, so it is the perimeter of a rectangle $12$ m by $8$ m:

$P = 2(l + w) = 2(12 + 8) = 40.$

The plot is $12$ m long and $8$ m wide, and it takes $40$ m of fencing to enclose.

The inner fence runs from the bottom side to the top side, so it spans the full height of the plot and is the same length as the two vertical sides, $8$ m. Nothing new needs measuring for it.

Each bed is itself a rectangle, and each still reaches the full height, so each is $8$ m from bottom to top. What separates them is their widths, which are the two gaps the fence line leaves. The bed on the left of the fence runs from $x = -10$ to $x = -6$, and the bed on its right from $x = -6$ to $x = 2$:

$|-6 - (-10)| = 4, \qquad |2 - (-6)| = 8.$

Multiply each width by the shared height to get the two areas:

$A_{\text{left}} = 4 \times 8 = 32, \qquad A_{\text{right}} = 8 \times 8 = 64.$

Now check them against the plot. The whole plot is $12$ m by $8$ m, so its area is $96$ square meters, and the two beds come to $32 + 64 = 96$ square meters as well. That agreement is worth running every time a figure is cut in two: the pieces must account for the whole, and if they do not, one of the widths has been read off wrongly.

Both beds run the full height of the plot, so both are $8$ m from bottom to top, and each area is that shared height times the bed's own width. When two rectangles have the same height, then, comparing their areas is the same as comparing their widths, and nothing else about them matters.

The fence at $x = -6$ leaves widths of $4$ m and $8$ m. Those are not equal, so the areas are not equal either, and the gardener is wrong:

$4 \times 8 = 32, \qquad 8 \times 8 = 64.$

The same reasoning says where the fence would have to go. For the two beds to match, each must take half the plot's width, and the plot is $12$ m across, so each bed must be $6$ m wide. Starting from the left side of the plot at $x = -10$ and moving $6$ m to the right lands on

$-10 + 6 = -4.$

Check that line from both directions before trusting it. From $x = -10$ to $x = -4$ is $6$ m, and from $x = -4$ to $x = 2$ is $6$ m as well, so each bed would be $6$ m by $8$ m, an area of $48$ square meters apiece, and the two together still come to the $96$ square meters of the plot.

One caution about the argument. Comparing widths alone settled this only because the two beds have the same height. Cut a plot with a fence running the other way and the widths would be equal while the heights differed, so it is the shared dimension, not the widths as such, that makes the comparison legitimate.

Keep the two verdicts apart for each point: what the shortcut says, and where the point is. Comparing them only afterwards is what makes this a test of the shortcut rather than a restatement of it.

The point $(12, 5)$ carries no minus sign, so the shortcut says Quadrant I. Its first coordinate is positive, putting it right of the y-axis, and its second is positive, putting it above the x-axis, so it is in the upper right. The shortcut is right.

The point $(-8, 3)$ carries one minus sign, so the shortcut says Quadrant II. It is $8$ units left of the y-axis and $3$ units above the x-axis, the upper left. The shortcut is right again.

The point $(7, -8)$ also carries one minus sign, so the shortcut says Quadrant II again. But this point is $7$ units right of the y-axis and $8$ units below the x-axis, which is the lower right:

$(7, -8): (+, -) \Rightarrow \text{Quadrant IV}.$

So the first failure is at the third point, and the two successes before it are worth noticing rather than skipping. A rule that has just worked twice is exactly the kind a student keeps, which is why a shortcut has to be tested on cases that could break it and not only on the ones that come to hand first.

Go back to what each coordinate is for. The first records the sideways position, measured from the y-axis, so its sign says which side of the y-axis the point is on. The second records the height, measured from the x-axis, so its sign says which side of the x-axis the point is on.

A quadrant is nothing more than a choice of one side of each axis: right or left, together with above or below. That is why there are four of them, and why the ordered pair of signs names one exactly:

$\text{I } (+,+), \quad \text{II } (-,+), \quad \text{III } (-,-), \quad \text{IV } (+,-).$

Counting destroys the very thing that distinguishes those four labels. A count returns only how many minus signs there are, never which coordinate carried one, so any two pairs with the same number of minus signs must receive the same verdict from it. Two of the points in the stem are such a pair:

$(-8, 3) \Rightarrow \text{II}, \qquad (7, -8) \Rightarrow \text{IV}.$

Each has exactly one minus sign, and they sit in different quadrants, so no rule that only counts can separate them. The signs are not a tally; they are two answers to two different questions, and the order they are written in is what says which question each is answering.

Start from what the shortcut is capable of saying. Its three verdicts are Quadrant I, Quadrant II and Quadrant III, so whatever it is given, one of those three comes back.

Now take the quadrants one at a time and compare. A point of Quadrant I has both coordinates positive, so no minus sign, and the shortcut answers I: correct. A point of Quadrant II has a negative x-coordinate and a positive y-coordinate, so exactly one minus sign, and the shortcut answers II: correct. A point of Quadrant III has both negative, so two minus signs, and the shortcut answers III: correct. A point of Quadrant IV has a positive x-coordinate and a negative y-coordinate, so exactly one minus sign again, and the shortcut answers II:

$\text{II } (-,+), \qquad \text{IV } (+,-).$

Those two patterns carry one minus sign apiece, so a rule that only counts is bound to give them the same verdict. The whole of Quadrant IV is therefore misnamed, every point of it, and not by accident: the shortcut's list of possible verdicts never contained that quadrant in the first place.

The points on an axis are the second family, and they fail for a different reason. A point on an axis has a zero coordinate, and $0$ is written with no minus sign, so such a point is fed into the shortcut like any other and comes back with a quadrant attached. But a point on an axis is in no quadrant at all, since the quadrants are the regions strictly between the axes. Any rule that always answers with a quadrant is therefore wrong about every one of these points, wherever they lie and whichever axis they are on.

Put together: the shortcut is right on Quadrants I, II and III, and wrong on Quadrant IV and on both axes. Notice the shape of that conclusion. Testing it on more points from the first three quadrants could never have found either family, which is the usual reason a broken shortcut survives so long.

A mirror image sits directly across the mirror line from the original, the same distance away, so the direction the mirror runs decides which coordinate can move.

The x-axis is horizontal, so crossing it is a straight up-or-down move. The point stays in its own column, which keeps the x-coordinate at $6$, and it lands the same distance on the far side of the axis, which turns the height $-9$ into $9$:

$W = (6, -9) \to W_1 = (6, 9).$

The y-axis is vertical, so crossing that one is a straight sideways move. Now the height is what survives untouched, staying at $-9$, while the sideways position flips to the other side of the axis:

$W \to W_2 = (-6, -9).$

Each of these two reflections changed exactly one sign, and each time it was the sign of the coordinate measured across the mirror: crossing the horizontal x-axis changed the height, while crossing the vertical y-axis changed the sideways position.

The reflection left the x-coordinate alone, so $W$ and $W_1$ both stand in the column $x = 6$. That shared coordinate is what puts the pair within reach: the segment joining them is vertical, so the y-coordinates are the ones that differ and their absolute difference is the length.

$d = |9 - (-9)| = |9 + 9| = 18.$

Now compare that with how far $W$ is from the mirror itself. $W$ sits at the height $-9$, which is $9$ units below the x-axis, and the reflection puts its image $9$ units above. The gap between them is those two stretches laid end to end, which is why it comes to twice $9$ rather than to $9$ or to anything else:

$18 = 2 \times 9.$

That doubling is not special to this point. A reflection always lands the image the same distance on the far side of the mirror, so a point and its image are always twice as far apart as the point is from the axis it was reflected across.

A mirror image is the point directly across the mirror from the original and the same distance from it. Directly across a horizontal mirror means straight up or straight down, since that is the direction at right angles to it, so the whole journey from a point to its image is vertical.

A vertical journey changes no sideways position. The image therefore stands in the same column as the point, and its x-coordinate is the point's x-coordinate, untouched. That is the first half, and notice it never mentioned any particular point.

For the second half, the distance from a point at height $y$ to the x-axis is $|y|$, and the image is that same distance away on the opposite side. Being on the opposite side and equally far from zero is exactly what the opposite number is, so the height $y$ comes back as $-y$:

$(x, y) \to (x, -y).$

Now ask which points are left where they were. Such a point must satisfy $-y = y$, since its x-coordinate never moved and only its height is in question, and the only number equal to its own opposite is $0$. So the unmoved points are exactly those with $y = 0$, which is to say the points of the x-axis itself, and that is no surprise: a point standing on the mirror is already its own image, at a distance of nothing from it.

The reasoning runs both ways, which is what makes exactly the right word here. Every point on the x-axis is unmoved, and every unmoved point is on the x-axis, so no other point anywhere in the plane survives the reflection unchanged.