Points, Lines, and Angles: Free Response

Each pair turns on a different feature, so take them one at a time.

A segment is the straight piece between two points, and both of its ends are endpoints. Naming them in the other order describes the same piece, so the first pair names one object.

A ray has a single endpoint, and the convention is to name that endpoint first. So $\overrightarrow{QP}$ starts at $Q$ and runs through $P$, while $\overrightarrow{PQ}$ starts at $P$ and runs through $Q$. They start at different points and run in opposite directions, so they are two different objects. They do overlap: every point between $P$ and $Q$ lies on both of them, which is why the drawing alone will not separate them.

A line has no endpoints at all, so the only question is which points it passes through. Exactly one line passes through two given points, and $P$, $Q$ and $R$ all lie on one line, so the two names in the third pair pick out that same line.

$\overline{PR} = \overline{RP}, \qquad \overrightarrow{QP} \neq \overrightarrow{PQ}, \qquad \overleftrightarrow{PQ} = \overleftrightarrow{QR}.$

The student's sentence hides a jump: it takes a failure for three points and reads it back onto pairs. Ask first what collinearity can ever rule out.

Any two points already have a line through them, and there is exactly one such line. So a pair of points is collinear no matter which two points are chosen, and saying so about a pair carries no information at all. Three is the first number of points for which the question is worth asking, because a third point can miss the line the first two determine. That is what happens with $Q$, $R$ and $S$: the line through $Q$ and $R$ is the horizontal one, and $S$ does not lie on it.

Now test each triple. The points $P$, $Q$ and $R$ are stated to lie on one line. Every other triple contains $S$ together with two of the first three, and those two determine the horizontal line, which $S$ misses.

$\{P, Q, R\} \;\text{collinear}; \qquad \{P, Q, S\}, \; \{P, R, S\}, \; \{Q, R, S\} \;\text{not}.$

Suppose, for the sake of argument, that the two lines met at two different points, and call those points $X$ and $Y$. Each of the two lines then passes through $X$ and through $Y$.

That is one claim too many. Exactly one straight line passes through $X$ and $Y$, so the two lines are not two lines at all: they are the same line, drawn twice.

$\ell = \overleftrightarrow{XY} \quad \text{and} \quad m = \overleftrightarrow{XY} \;\Rightarrow\; \ell = m.$

So the supposition is impossible for lines that really are different, and two different lines share at most one point. Notice what the argument does not claim: it does not say two lines must meet. Parallel lines share no point at all, and the result rules out two or more shared points, leaving exactly the two cases we can draw, one shared point or none.

The two marked angles sit next to each other at $B$: they share the side $\overrightarrow{BD}$ and do not overlap, so they are adjacent and their measures add to the measure of the angle from $\overrightarrow{BA}$ round to $\overrightarrow{BC}$.

$\angle ABC = \angle ABD + \angle DBC = 23^\circ + 67^\circ = 90^\circ$

The second route never touches those two measures. The angles $\angle ABC$ and $\angle CBE$ share the side $\overrightarrow{BC}$, and their outer sides $\overrightarrow{BA}$ and $\overrightarrow{BE}$ point in opposite directions along one straight line, so the two form a linear pair and add to a straight angle.

$\angle ABC = 180^\circ - 90^\circ = 90^\circ$

The two routes use different given information and agree, which is the check worth running: if the figure had been marked inconsistently, these two answers would have disagreed.

The angles $\angle ABD$ and $\angle DBE$ share the side $\overrightarrow{BD}$, and their outer sides run opposite ways along the straight line through $A$, $B$ and $E$. So they are a linear pair and add to a straight angle.

$\angle DBE = 180^\circ - 23^\circ = 157^\circ$

The figure offers a second route as a check: $\angle DBE$ is made of the adjacent angles $\angle DBC$ and $\angle CBE$, giving $67^\circ + 90^\circ$, the same total.

Now sort the three measures against the two landmarks. Both $23^\circ$ and $67^\circ$ fall short of a right angle, so those two angles are acute. The third sits between a right angle and a straight angle, since $90^\circ < 157^\circ < 180^\circ$, so it is obtuse. Nothing here is reflex: a reflex angle would have to be more than a straight angle, and the largest angle marked in the figure stops short of that.

A single-letter name works only when exactly one angle sits at that vertex, because the letter carries nothing but the corner. Here four rays leave $B$, namely $\overrightarrow{BA}$, $\overrightarrow{BD}$, $\overrightarrow{BC}$ and $\overrightarrow{BE}$, and any two of them bound an angle: $\angle ABD$, $\angle ABC$, $\angle ABE$, $\angle DBC$, $\angle DBE$ and $\angle CBE$. The name $\angle B$ would have to pick one of those six, and it gives no way to.

The three-letter form fixes that by spending its middle letter on the vertex and its outer letters on a point along each side. So $\angle ABC$ and $\angle CBA$ both have vertex $B$, and both name the sides through $A$ and through $C$; swapping the outer letters swaps nothing but the order in which the two sides are mentioned, so the two names describe one angle. In $\angle BAC$ the middle letter is $A$, so the vertex has moved: it names the angle at $A$ between $\overrightarrow{AB}$ and $\overrightarrow{AC}$, a different angle at a different vertex.

$\angle ABC = \angle CBA \neq \angle BAC$

Both of these angles sit at the upper crossing, so only the two lines through that crossing matter.

The angle $a$ and the marked angle share the part of the rail running to the right of the crossing, and their outer sides are the two directions of the brace, which together make one straight line. So they are a linear pair and add to a straight angle.

$a = 180^\circ - 116^\circ = 64^\circ$

The angle $b$ sits directly across the crossing from the marked angle, sharing only the vertex, so the two are vertical angles and are equal.

$b = 116^\circ$

Neither step mentioned the lower rail. A linear pair needs one straight line through the vertex, and a vertical pair needs the two lines that cross there, so both facts hold at any crossing whatsoever. Had the rails been drawn at different tilts, these two answers would have been exactly the same.

These two angles sit at the lower crossing, so reaching them means travelling from one crossing to the other, and that is the step the chevrons pay for.

The angle $d$ occupies the same position at its crossing as the marked angle does at its own: below the rail and to the right of the brace. Corresponding angles at a transversal cutting parallel lines are equal.

$d = 116^\circ$

The angle $c$ lies between the rails on the same side of the brace as the marked angle, so the two are co-interior and add to a straight angle.

$c = 180^\circ - 116^\circ = 64^\circ$

A second route to $c$ is worth noticing: $c$ and $d$ are a linear pair along the lower rail, so $c = 180^\circ - d$ gives the same value.

If the rails were not parallel, both routes would collapse. Corresponding and co-interior angles are relationships between two different crossings, and nothing ties the two crossings together except the rails running in the same direction. The measures $c$ and $d$ would then be free to be almost anything, and the marked angle at the upper crossing would say nothing at all about them.

Write $u$ for the interior angle at the upper crossing on one side of the brace, and $v$ for the interior angle at the lower crossing on that same side. The two sit at different crossings, so the argument needs one step that travels between them and one step that stays put.

For the travelling step, look at the angle at the lower crossing that occupies the same position as $u$ does at its own crossing: below its rail, on the same side of the brace. Corresponding angles are equal when the rails are parallel, so that angle also measures $u$.

For the second step, stay at the lower crossing. That corresponding angle and $v$ share the part of the lower rail on their side of the brace, and their outer sides are the two directions of the brace, so they are a linear pair.

$u + v = 180^\circ$

Now read the types off that total. If $u$ is less than a right angle, then $v = 180^\circ - u$ is more than a right angle, so one is acute and the other obtuse. If $u$ is more than a right angle the same sentence applies with the two swapped. The only remaining possibility is that $u$ is exactly a right angle, and then so is $v$. So the pair is either two right angles or one acute angle with one obtuse angle, and no other combination can occur: two acute angles would fall short of $180^\circ$ and two obtuse angles would overshoot it.

That middle case has a name in the figure. Both angles are right angles exactly when the brace meets the rails at a right angle, that is, when the brace is perpendicular to them; and if the brace is perpendicular to one of two parallel rails, the corresponding angle makes it perpendicular to the other as well.

Neighbouring angles at the crossing are exactly the ones a linear pair applies to: two angles that sit next to each other there share a side, and their outer sides are the two halves of one of the lines, so they complete a straight angle.

Take the neighbours in order. From $\angle 1$ and $\angle 2$,

$\angle 2 = 180 - x.$

Now $\angle 2$ and $\angle 3$ are neighbours along the other line, so the same rule applies to them, and the two subtractions undo each other:

$\angle 3 = 180 - (180 - x) = x.$

One more step, using $\angle 3$ and $\angle 4$, closes the circuit:

$\angle 4 = 180 - \angle 3 = 180 - x.$

So one measurement really does settle the crossing: every angle there is either $x$ or $180 - x$, and which of the two it is alternates as you go round. Adding the four measures, the $x$ terms cancel in pairs:

$x + (180 - x) + x + (180 - x) = 360.$

The total is the same number whatever $x$ was, which is the angles-around-a-point fact turning up on its own rather than being assumed.

Tilting the lines changes the measure of $\angle 1$, and nothing else is free, because the other three angles are forced by it. Write them out and count how many different expressions there actually are.

$\angle 1 = x, \quad \angle 2 = 180 - x, \quad \angle 3 = x, \quad \angle 4 = 180 - x$

There are four angles but only two expressions, so at most two different measures can appear no matter how the lines are tilted. Four different measures is impossible, and so is three.

That leaves the question of when there are two and when fewer. The two expressions give the same value only when

$x = 180 - x,$

and the value that satisfies this is a right angle, which is the case where the two lines are perpendicular. Then all four angles are right angles and there is a single measure. For any other tilt the two expressions differ, so exactly two different measures appear, each of them twice.

So the student has the picture upside down: tilting the lines further does not spread the four measures out, it only moves the single value $x$, and the crossing keeps its two-and-two pattern throughout.

A claim about equal angles is easiest to break where as many angles as possible are equal, so look for the crossing that makes the measures collapse together.

That crossing is the perpendicular one. Taking $x$ to be a right angle,

$\angle 1 = \angle 2 = \angle 3 = \angle 4 = 90^\circ,$

since each angle is either $x$ or $180 - x$, and those are the same value here. This is a perfectly ordinary crossing of two straight lines, so it is a case the claim has to cover.

Now point at the failure. The angles $\angle 1$ and $\angle 2$ have equal measures, and they are neighbours: they share a side, which is exactly what an opposite pair does not do. So two equal angles at this crossing are not an opposite pair, and the claim is false.

It is worth saying how narrowly it fails. At any crossing that is not perpendicular the two expressions $x$ and $180 - x$ are different, so the only equal pairs are the opposite ones and the claim holds there. The perpendicular crossing is the single case that breaks it, which is why a picture of a tilted crossing will never reveal the fault.

Take the steps in order, because a step can only be called the first fault once the ones above it have been cleared.

Step 1 compares the marked angle with the angle across the upper crossing from it. Those two are vertical angles, formed by the two straight lines $p$ and $n$ that meet there, so they are equal and the step is sound.

Step 2 compares the marked angle with its neighbour along $p$. They share a side, and their outer sides are the two halves of $n$, so they are a linear pair and complete a straight angle.

$180^\circ - 74^\circ = 106^\circ$

That step is sound as well, and both of the sound steps have something in common: each uses only the two lines through the upper crossing.

Step 3 is different. It compares an angle at the upper crossing with an angle at the lower one, and the corresponding-angle relationship it appeals to holds only when the two crossed lines are parallel. The figure carries no chevrons and the question states nothing about $p$ and $q$ beyond their both being crossed by $n$, so that relationship has not been earned. It is the first unjustified step, and the information it needs is precisely a statement or marking that $p$ and $q$ are parallel.

Sort the figure into what is marked and what is merely drawn. Marked: the two lines $p$ and $n$ cross, the angle between them at that crossing is $74^\circ$, the line $q$ crosses $n$, and $x$ is the angle in the matching corner there. Nothing is marked about the direction of $q$.

Now test whether $x$ is forced, by trying to change it while keeping every marked fact. Hold $p$ and $n$ exactly where they are and swing $q$ about the point where it meets $n$. The $74^\circ$ angle is untouched, since $q$ is not one of the lines that make it. The lower crossing is still a crossing of $q$ with $n$, and $x$ is still the angle in the matching corner. Yet $x$ has changed. Swinging $q$ far enough makes $x$ take any value strictly between a zero angle and a straight angle, since only those two extremes would lay $q$ along $n$ and destroy the crossing:

$0^\circ < x < 180^\circ$

So the marked information does not fix $x$ at all. One value it could take is $87^\circ$, and that reading contradicts nothing: the angle at the upper crossing is still $74^\circ$, the two crossings are still crossings, and no marking claims any relationship between them.

The student's $74^\circ$ is not wrong so much as unearned. It is one of the values $x$ could take, the one it takes when $p$ and $q$ happen to be parallel, and the figure gives no reason to choose it over the rest.

Ask of each fact how much of the figure it has to look at.

A linear pair lives at one vertex: two angles sharing a side, with their outer sides completing one straight line through that vertex. Vertical angles live at one vertex too, needing the two straight lines that meet there and nothing else. Both facts can be checked with the rest of the figure covered up, so both hold at any crossing whatsoever, however the lines are tilted. At the upper crossing they settle all four angles between them:

$74^\circ, \quad 106^\circ, \quad 74^\circ, \quad 106^\circ$

The matching-position fact is not like that. It compares an angle at one crossing with an angle at another, and the two crossings are separate pieces of the figure. Something has to tie them together, and the only thing available is the two crossed lines running in the same direction: slide one crossing along the transversal and it lands exactly on the other, carrying each angle onto its match, but only if the lines are parallel. The same goes for the alternate interior and co-interior facts, which are the matching-position fact with a vertical pair or a linear pair applied afterwards.

So the rule to carry to a new figure is a question about what a fact compares. If both angles sit at the same crossing, the two lines meeting there settle them, so a vertical pair or a linear pair is available with no further marking. If they sit at different crossings, first find the marking or the statement that says the crossed lines are parallel, and if there is none, the relationship is not available.