Points, lines and planes
Every definition is made of other words. Where does geometry’s chain of definitions stop?
Where you are 8 in Points, lines and planes
Why this exists
Look up “point” in a dictionary and you might read “an exact position”. Look up “position”: “the place where something is”. Look up “place”: “a particular point or area”. Three steps, and you are back where you started.
Every dictionary does this. A definition explains a word with other words, and with a finite stock of words the chain can only loop back or stop at words that are never defined. A loop is fine for everyday speech, since you learned “place” by being in places, but useless for mathematics, where every claim has to be traced back to something. So geometry stops the chain on purpose: it picks a few words, declines to define them, and builds everything else on them. Those words are the undefined terms, and in school geometry there are three: point, line and plane.
The first words defined from them say where points sit. Mark three dots on a page and try to rule one straight line through all of them. Two dots always allow it, but the third has to sit in exactly the right place, and a dot placed at random almost never does. Points that do line up are called collinear. The same question comes up one size larger: a flat plane can always be laid through three points, but a fourth point usually sticks out above or below it. The lucky case there is coplanar.
Whether points line up decides what they can build. Three noncollinear points are the corners of a triangle; four noncoplanar points are the corners of a solid. Many later definitions, from the plane ABC to skew lines, have one of these words inside them.
Definition
The three undefined terms
An undefined term is a word accepted as a starting point without a definition. It is given a description, which tells you what to picture, and it is governed by postulates, the rules assumed about it. The description is for your intuition. The postulates are what proofs use.
Point. A location, with no length, width or thickness. It is named by a capital letter, as in point A, and drawn as a dot, but the dot is only a marker: the point has no size at all.
Line. Perfectly straight, with no thickness, going on without end in both directions and containing infinitely many points. It is named by any two of its points under a double arrow, \overleftrightarrow{AB}, or by one lowercase letter such as \ell or m, and drawn with an arrowhead at each end to show that it does not stop. In geometry a line is always straight.
Plane. A perfectly flat surface with no thickness, going on without end in every direction along it. It is named by one script capital such as \mathcal{P}, or by three of its points that do not all lie on one line, as in plane ABC. It is drawn as a parallelogram, one piece of it seen at a slant.
Everything else is defined from these. The first defined words, collinear and coplanar, follow next; a segment is two points on a line and every point between them. Example 6 unpacks one definition all the way down.
Collinear and coplanar points
Points are collinear when one line passes through all of them. Points that are not collinear are noncollinear.
Points are coplanar when one plane contains all of them. Points that are not coplanar are noncoplanar.
In Figure 2, A, B and C are collinear: the line \ell passes through all three. A, B and D are noncollinear, since no single line reaches all three, but they are coplanar, because plane \mathcal{M} holds them. In fact A, B, C and D are coplanar, all four sitting in \mathcal{M}. The point E is above the plane, so A, B, D and E are noncoplanar.
The words ask for one line, or one plane, holding every point at once. The same words are used for lines: two lines are coplanar when one plane contains both of them.
Intuition
Models of the three terms
A pencil tip or a star in the night sky is a model of a point, a stretched string or the edge of a ruler a model of a line, a tabletop or a still pond a model of a plane. Each has some size. The geometric object is what you get by shrinking the thickness to nothing and pushing the edges out forever.
David Hilbert, who rebuilt geometry on undefined terms in 1899, is reported by his student and biographer Otto Blumenthal to have remarked that one must always be able to say “tables, chairs and beer mugs” in place of “points, lines and planes”. A proof uses only the rules, so it must survive any change of names. Pictures help you find proofs; they are never part of one.
Strings, glass and stools
Stretch a string tight between two nails. Any two nails can be joined that way, which is why two points are always collinear. A third nail is collinear with them only if it touches the string.
Now rest a sheet of glass on the tips of three upright pins. Whatever their heights, the glass settles: three points always have a plane through them. That is why a three-legged stool stands steady on any floor, while a four-legged table can rock: the four spots of floor under its feet need not be coplanar.
Results
Where definitions stop
Every chain of definitions ends at undefined terms. In geometry it ends at point, line and plane, with relation words such as “on” and “between”.
An undefined term is used in a proof only through its postulates. The description can suggest a fact. Only a postulate, or a theorem proved from the postulates, can establish it.
Four of the first postulates, as examples:
- Through any two different points there is exactly one line.
- Every line contains at least two points.
- Through any three noncollinear points there is exactly one plane.
- If two different points of a line lie in a plane, the whole line lies in that plane.
What follows about how lines and planes meet is in lines and planes in space.
Collinear and coplanar points
Two points. By the first postulate, any two points are collinear.
Three points. Any three points are coplanar. When the three are noncollinear, exactly one plane contains them. That is why a plane can be named by three of its points, as in plane ABD.
A line and a plane. By the fourth postulate, a plane holding two different points of a line holds the whole line. In particular, collinear points are always coplanar.
Testing three points for collinearity
The slope test. Three different points A, B, C on the coordinate plane, with x_A \ne x_B and x_A \ne x_C, are collinear exactly when the slope from A to B equals the slope from A to C:
When two of the points share an x-coordinate, the three are collinear exactly when all three share it.
The distance test. Three different points are collinear exactly when the longest of the three distances between them equals the sum of the other two, and then the point that is not an end of the longest distance lies between the other two. With the points named so that AC is the longest,
holds exactly when B lies on \overline{AC}; otherwise AB + BC > AC.
Why it is true
Why some words must go undefined
That some words must go undefined is a counting argument.
Suppose every word in a system were defined. A word w_1 is defined using some word w_2 (not counting connecting words like “the”), w_2 using some w_3, and so on: the chain w_1, w_2, w_3, \dots never ends. With finitely many words, an endless list must repeat, so w_j = w_i for some earlier i, and w_i is defined, through the steps between, in terms of itself. A circular definition explains nothing to someone who does not already know the word. So a finite system of definitions with no circles must leave some word undefined.
Euclid, around 300 BCE, did try to define the three: “a point is that which has no part”, “a line is breadthless length”. His proofs never use these sentences, which lean on “part” and “breadth”, no easier than “point”. Modern geometry admits openly what his practice already showed.
Why any three points are coplanar
The postulates in Results are accepted without proof. The fourth, a line that stays in a plane, is what “flat” means.
Any three points are coplanar follows from them. If the three are noncollinear, the plane postulate hands you a plane. If they are collinear, they lie on a line \ell. Pick two different points X and Y of \ell (every line has at least two) and a point P off \ell. Then X, Y and P are noncollinear, so one plane holds them; that plane contains two different points of \ell, so it contains all of \ell, and with it all three points. The same argument shows that collinear points are always coplanar.
Why the slope test works
Suppose the slope from A to B and the slope from A to C are the same number m. The line through A and B passes through A with slope m, and so does the line through A and C. A non-vertical line is fixed by one point and its slope, so these are the same line, and it holds all three points. Going the other way, if all three lie on one non-vertical line, every pair of them gives the slope of that line, as the slope formula shows. The shared point A matters: equal slopes with no common point can mean parallel lines.
Why the distance test works
There are three places B can sit relative to A and C.
If B is on \overline{AC}, the two pieces make up the whole, AB + BC = AC, by the segment addition postulate. If B is off the line through A and C, then A, B, C are the corners of a triangle, and by the triangle inequality the bent path is strictly longer: AB + BC > AC. If B is on the line but beyond C, then C is between A and B, so AB = AC + CB is already longer than AC, and AB + BC > AC again; beyond A works the same way. So the sum equals AC only when B lies on the segment.
That settles the longest distance too. Of three different collinear points, one lies between the other two, and the distance across the outer pair is the sum of the two pieces, so it is the longest. The test uses distance alone, with no slopes and no coordinates.
Worked examples
Choosing the right undefined term
Say which undefined term each object is a model of, and one way it falls short: (a) a city on a map of the world, (b) a laser beam, (c) the surface of a frozen lake.
Solution(a) A point: the map marks only where the city is. But the city, and its dot, has a size.
(b) A line: the beam is straight and long for its width. But it has width, and it starts at the laser.
(c) A plane: the ice is flat and thin for how far it spreads. But it has thickness, a shore, and the curve of the Earth.
Naming a line
Figure 4 shows line m with points A, B and C on it. Write every name the line has.
A line can be named by any two of its points with the double arrow. There are three pairs: A and B, A and C, B and C. Each pair can be written in either order, since a line has no direction, so \overleftrightarrow{AB} and \overleftrightarrow{BA} are the same line. That gives 3 \times 2 = 6 two-letter names:
With the single letter m, that is 6 + 1 = 7 names, all for the same line.
Naming a plane
Figure 5 shows plane \mathcal{R} with four points W, X, Y and Z in it, no three of them on one line. How many names does the plane have, counting \mathcal{R} and every name made of three of these points?
A three-letter name needs three points of the plane not all on one line, and here every choice of three qualifies. Choosing three of the four is the same as leaving one out, which can be done in 4 ways: WXY, WXZ, WYZ and XYZ. Each choice can be written in 3 \times 2 \times 1 = 6 orders, so there are
three-letter names, such as plane WXY, plane YWX and plane ZXW. With \mathcal{R} as well, the plane has 25 names.
Reading collinear and coplanar points from a figure
Use Figure 2. Are A, B and D collinear? Are they coplanar? Are A, C, D and E coplanar?
SolutionOnly one line passes through A and B, and it is \ell. D is not on \ell, so no line holds A, B and D together: they are noncollinear. All three lie in plane \mathcal{M}, so they are coplanar.
For the four points, start with three of them. A, C and D are noncollinear, since D is off \ell, the only line through A and C. So exactly one plane holds A, C and D, and that plane is \mathcal{M}. E is not in \mathcal{M}, so no plane holds all four: A, C, D and E are noncoplanar.
Coplanar points on a box
Figure 6 shows a box. Are A, B, C, F coplanar? Are A, B, G, H coplanar?
A, B and C are three corners of the bottom face, and they are noncollinear, so exactly one plane holds them: the plane of the bottom face. F sits directly above B, off the bottom face, so it is not in that plane. A, B, C, F are noncoplanar.
For A, B, G, H look at the edges \overline{AB}, along the front of the bottom, and \overline{HG}, along the back of the top. Both run in the direction of \overline{DC}, so they are parallel, and parallel lines lie in one plane (two lines that lie in no common plane are skew). That plane holds all four points, so A, B, G, H are coplanar: the corners of the shaded slice.
Unpacking a definition
A triangle ABC is three points A, B and C that are not collinear, together with the segments \overline{AB}, \overline{BC} and \overline{CA}. Unpack this definition until only undefined terms and relation words are left.
Solution“Not collinear” is defined: the three points do not all lie on one line. That is down to point, line and “on”.
“Segment \overline{AB}” is defined: the points A and B and every point of \overleftrightarrow{AB} between them. That is down to point, line and “between”.
Put back together: a triangle is three points, not all on one line, with, for each two of them, those two points and every point between them on the line through them. Every word left is point, line, “on”, “between” or plain logic, exactly where the first result says the chain stops.
Testing a proposed definition
Someone suggests: “A line is the shortest path between two points.” Is this a good definition of a line?
SolutionNo, for two reasons. It describes the wrong object: the shortest path between two points stops at them, but a line goes on forever both ways. The shortest path is a segment, a piece of a line.
And it uses words harder than “line”. “Path” needs defining, and “shortest” needs length, which is measured along a straight path, that is, along a line. The sentence is true of lines once you have them, but it cannot be where they come from.
The slope test for points that line up
Are P(-2, -1), Q(1, 1) and R(7, 5) collinear?
SolutionMeasure both slopes from the shared point P. From P to Q:
From P to R:
The slopes are equal and both start at P, so one line through P carries both Q and R. The three points are collinear.
Check. The slope from Q to R should match too: \displaystyle \frac{5 - 1}{7 - 1} = \frac{4}{6} = \frac{2}{3}.
The slope test for points that nearly line up
Are A(-3, 7), B(1, 2) and C(5, -2) collinear?
SolutionFrom A to B:
From A to C:
The slopes differ, so the points are not collinear. The miss is small: the line through A and B drops 5 for every 4 across, so at x = 5 it reaches y = 2 - 5 = -3, and C sits one unit above that: easy to miss on a sketch, which is why arithmetic settles it.
Finding a coordinate that makes points collinear
Find k so that (1, -1), (4, 5) and (k, 13) are collinear.
SolutionThe first two points fix the line. Its slope is
For (k, 13) to be on that line, the slope from (1, -1) to it must also be 2:
Check. From (4, 5) to (8, 13) the slope is \displaystyle \frac{8}{4} = 2, the same again.
The distance test
(a) AB = 5, BC = 8 and AC = 13. Are A, B, C collinear?
(b) PQ = 7, QR = 11 and PR = 15. Are P, Q, R collinear?
Solution(a) The longest distance is AC = 13, and the other two add to 5 + 8 = 13. The longest equals the sum of the others, so the points are collinear, and B, the point not at either end of \overline{AC}, lies between A and C.
(b) The longest distance is PR = 15, but the other two add to 7 + 11 = 18. The path from P to Q to R is longer than the straight way from P to R, so it bends at Q. The points are not collinear; they are the corners of a triangle.
Always compare against the longest distance. Any other pairing adds the longest distance to something positive, so it can never equal the third.
Counting lines through points
Four different points A, B, C, D are given. How many lines pass through two of them (a) when no three of the points are collinear, (b) when A, B, C are collinear and D is not on their line?
Solution(a) Each pair of points has exactly one line through it. No line holds three of the points, so no line is counted by two different pairs, and the count of lines is the count of pairs: AB, AC, AD, BC, BD, CD, which is 6 lines.
(b) The pairs AB, AC and BC all give the same line, the one through A, B and C. The pairs AD, BD and CD give three more lines, new because they pass through D, and different from each other: a line through D and two of A, B, C would be their line, which misses D. That is 1 + 3, or 4 lines.
The same six pairs give different counts, so a count means nothing until its hypothesis is stated. Planes are counted the same way in lines and planes in space.
A geometry with only three points
Because “point” and “line” are undefined, they can stand for anything that obeys the rules. Take three rules:
- Rule 1. Through any two different points there is exactly one line.
- Rule 2. Every line contains at least two points.
- Rule 3. There are three points that do not all lie on one line.
Let the “points” be the letters A, B and C, the “lines” the pairs \{A, B\}, \{B, C\} and \{A, C\}, and “on” mean “is a member of”. Check the three rules, then use the rules alone to show that every point lies on at least two lines.
SolutionEach pair of points is itself a line, and the only one holding both (Rule 1); every line has two points (Rule 2); no line holds all three (Rule 3).
Now the proof, from the rules and nothing else. Take any point P. Rule 3 gives three points, so some point Q differs from P, and by Rule 1 there is a line \overleftrightarrow{PQ}. Rule 3 also says not every point is on \overleftrightarrow{PQ}; call a point off it R. By Rule 1 there is a line \overleftrightarrow{PR}, and it differs from \overleftrightarrow{PQ} because R is on one and not the other. So P lies on at least two lines.
In the model, A lies on \{A, B\} and \{A, C\}. But the proof never mentioned letters or straightness, so it holds in the ordinary plane too, and in anything that obeys the rules.
Boundaries
Undefined does not mean vague. The postulates pin down exactly how points, lines and planes behave; only what they are is left open, and no proof needs that. “A point has no size” tells you what to picture, and is never a reason in a proof.
The choice of three is a choice. Hilbert also left “lies on”, “between” and “congruent” undefined; other systems take only points as undefined and define a line as a set of points.
A model can obey some rules and break others. On a globe, call the great circles (circles, like the equator, centered at the Earth’s center) “lines”. Two points that are not opposite lie on exactly one, but opposite points such as the poles lie on infinitely many, so Rule 1 fails. A model that keeps Rule 1 but breaks the parallel postulate gives a whole different geometry.
Small counts say nothing. The words collinear and coplanar carry information only from three points on for collinear and from four points on for coplanar.
Straight means straight. Points on a curve need not be collinear: a line meets a circle in at most two points, so three different points on a circle are never collinear.
On a flat page, everything is coplanar. In plane geometry all points share the plane of the page, so “coplanar” only separates cases in space.
The tests need exact values. A drawing cannot prove points collinear, and neither can measured lengths: 5.0 + 8.0 = 13.0 on a ruler also fits three points that bend by less than the ruler shows.
Where this goes next
The first defined words, collinear and coplanar, say where points sit. Three noncollinear points make a triangle and fix a plane, and from there the first consequences of the postulates say how lines and planes meet, and what the solids of space are built from: lines and planes in space. The first flat figures are segments and rays and angles.
On the coordinate plane, three collinear points make a “triangle” of area zero, another test for lining up, met in the area of a polygon from its corners. The mirror idea is for lines: lines are concurrent when they share one point, and the special points of a triangle are all statements that three lines are concurrent.
Axioms, definitions and theorems looks at unproved starting points in all of mathematics and how the three kinds of statement work together, and converses and biconditional definitions shows why a good definition works in both directions.
Exercises
Describing and naming
- Say which undefined term each object is a model of: (a) the tip of a needle, (b) a tightly stretched wire, (c) a sheet of glass.
- Using Figure 4, give three different names for line m.
- Using Figure 5, give two different names for the plane.
- Using Figure 5, give two names for line n that use points.
- Using Figure 5, is \overleftrightarrow{WZ} the same line as n? Explain.
Counting names
- Four points P, Q, R and S are marked on line k. How many names does the line have, counting k and every name of the form \overleftrightarrow{XY}?
- A line has exactly 12 names of the form \overleftrightarrow{XY} using the points marked on it. How many points are marked?
- Five points lie in plane \mathcal{Q}, no three of them on one line. How many three-letter names can the plane be given using these points?
Collinear and coplanar in a figure
Use Figure 2 for Exercises 9 to 12.
- Name three collinear points.
- Name four coplanar points.
- Are A, D and E coplanar? Explain.
- Are B, C, D and E coplanar? Explain.
Points on a box
Use Figure 6 for Exercises 13 to 15.
- Are A, D, H and E coplanar?
- Are B, D, H and F coplanar? Explain.
- Are A, C, G and F coplanar? Explain.
Definitions
- Each sentence tries to define “point”. Name a word in each that is no easier than “point” itself: (a) a point is a dot, (b) a point is a location, (c) a point is where two lines cross.
- A dictionary gives: line, a long thin mark; mark, a trace left on a surface; trace, a line or mark left behind. In how many steps does the chain from “line” first come back to a word already used?
- Explain why “a plane is a flat surface” is a description and not a definition.
The slope test
Decide whether the three points are collinear.
- (0, 1), (2, 7), (5, 16)
- (-3, 4), (1, 2), (7, -1)
- (1, 1), (4, 3), (10, 6)
- (2, -1), (2, 5), (2, 11)
- Find k so that (1, 4), (3, 10) and (6, k) are collinear.
- Find every value of k that makes (k, 1), (3, k) and (5, 5) collinear.
The distance test
Decide whether the three points are collinear, and if they are, which one is between the other two.
- AB = 6, BC = 9, AC = 15
- PQ = 4, QR = 10, PR = 7
- XY = 12, YZ = 5, XZ = 7
- A(0, 0), B(6, 8), C(9, 12), using the distance formula.
- A, B and C are collinear, AB = 4 and BC = 6. Find every possible value of AC.
Counting lines
- Five different points are given, no three of them collinear. How many lines pass through two of them?
- Five different points are given. A, B and C lie on one line, and no other three of the five are collinear. How many lines pass through two of them?
Rules and models
Use Rules 1, 2 and 3 from Example 13.
- Let the points be A, B, C and D, and the lines every pair of them. How many lines are there, and do all three rules hold?
- Let the points be A, B, C and D, and the lines \{A, B, C\}, \{A, D\}, \{B, D\} and \{C, D\}. Do all three rules hold?
- Let the points be A, B, C and D, and the lines \{A, B\} and \{C, D\}. Which rule fails, and why?
- In the model of Exercise 32, how many lines pass through each point?
Reasoning
- Using only Rules 1, 2 and 3, prove that there are at least three different lines.
- Using only Rules 1, 2 and 3, prove that for every line there is a point not on it.
- Explain why a system in which every word is defined, using only finitely many words, must contain a circular definition.
- A dot drawn on paper is about half a millimeter wide. Does that make “a point has no size” false? Explain.
- A, B, C are collinear, B, C, D are collinear, and B and C are different points. Show that A, B, C, D are all collinear. Why does it matter that B \ne C?
- Explain why collinear points are always coplanar, but coplanar points need not be collinear. Give an example from Figure 2.
Answers
- (a) A point: it marks a location. (b) A line: it is straight and very long for its width. (c) A plane: it is flat and very thin for how far it spreads.
- Any three of m, \overleftrightarrow{AB}, \overleftrightarrow{BA}, \overleftrightarrow{AC}, \overleftrightarrow{CA}, \overleftrightarrow{BC}, \overleftrightarrow{CB}.
- Plane \mathcal{R}, and any three of the points in any order, such as plane WXY or plane XYZ.
- \overleftrightarrow{WX} and \overleftrightarrow{XW}, since W and X are the only marked points on n.
- No. Z is not on n, so a line through W and Z contains a point that n does not.
- 13 names. There are 4 \times 3 = 12 ordered pairs of different points, each a name such as \overleftrightarrow{PQ}, and k makes 12 + 1 = 13.
- 4 points. With n points there are n(n - 1) two-letter names, and n(n - 1) = 12 gives n = 4 (the other solution, n = -3, is not a number of points).
- 60 names. Three of five points can be chosen in 10 ways, each written in 6 orders: 10 \times 6 = 60.
- A, B and C. All three lie on the line \ell.
- A, B, C and D. All four lie in plane \mathcal{M}.
- Yes. Any three points are coplanar. The plane holding them is not \mathcal{M}, because E is not in \mathcal{M}, but some plane holds all three.
- No. B, C, D are noncollinear, since D is off \ell, the only line through B and C. So exactly one plane holds them, and it is \mathcal{M}. E is not in \mathcal{M}.
- Yes. They are the four corners of the left face of the box.
- Yes. \overline{BF} and \overline{DH} are both upright edges, so they are parallel, and parallel lines lie in one plane. That plane holds all four points.
- No. A, C, G are noncollinear, so one plane holds them: the upright slice ACGE standing on the diagonal \overline{AC} of the bottom face. F is directly above B, which is off that diagonal, so F is not in the slice.
- (a) “dot”, which has a size and is only a marker. (b) “location”, which a dictionary defines back in terms of points or places. (c) “line”, another undefined term.
- 3 steps: “line” to “mark” (step 1), “mark” to “trace” (step 2), and “trace” back to “line” and “mark” (step 3).
- It leans on “flat” and “surface”, and making “flat” precise needs straight lines lying in the surface, the very idea the plane was meant to supply. It tells you what to picture, nothing more.
- Collinear. The slope from (0, 1) to (2, 7) is \displaystyle \frac{6}{2} = 3, and from (0, 1) to (5, 16) it is \displaystyle \frac{15}{5} = 3.
- Collinear. From (-3, 4) the slopes are \displaystyle \frac{2 - 4}{1 + 3} = -\frac{1}{2} and \displaystyle \frac{-1 - 4}{7 + 3} = -\frac{5}{10} = -\frac{1}{2}.
- Not collinear. From (1, 1) the slopes are \displaystyle \frac{2}{3} and \displaystyle \frac{5}{9}, which differ.
- Collinear. All three have x-coordinate 2, so they lie on the vertical line x = 2. The slope formula does not apply here.
- k = 19. The slope of the first two is \displaystyle \frac{10 - 4}{3 - 1} = 3, so \displaystyle \frac{k - 4}{6 - 1} = 3, giving k - 4 = 15.
- \displaystyle k = 5 + 2\sqrt{2} or \displaystyle k = 5 - 2\sqrt{2}. The slope from (3, k) to (5, 5) is \displaystyle \frac{5 - k}{2} and from (k, 1) to (5, 5) it is \displaystyle \frac{4}{5 - k}. Setting them equal gives (5 - k)^2 = 8, so \displaystyle 5 - k = \pm 2\sqrt{2}, never zero, so both slopes exist.
- Collinear, with B between A and C. The longest is AC = 15 = 6 + 9.
- Not collinear. The longest is QR = 10, but 4 + 7 = 11.
- Collinear, with Z between X and Y. The longest is XY = 12 = 5 + 7.
- Collinear, with B between A and C. \displaystyle AB = \sqrt{36 + 64} = 10, \displaystyle BC = \sqrt{9 + 16} = 5 and \displaystyle AC = \sqrt{81 + 144} = 15, and 15 = 10 + 5.
- AC = 10 or AC = 2. If B is between, AC = 4 + 6 = 10. If A is between, BA + AC = BC, so AC = 6 - 4 = 2. If C were between, AC + CB = AB would give AC = 4 - 6 = -2, which is impossible.
- 10. No line holds three of the points, so each pair gives its own line, and there are \displaystyle \frac{5 \cdot 4}{2} = 10 pairs.
- 8. Of the 10 pairs, AB, AC and BC give one line between them, and no other line holds three of the points, so the other 7 pairs give 7 different lines: 1 + 7 = 8.
- 6 lines, and all three rules hold. There are \displaystyle \frac{4 \times 3}{2} = 6 pairs; each pair is exactly one line, each line has two points, and A, B, C are in no one line.
- Yes. Any two of A, B, C lie together only in \{A, B, C\}, and each pair with D is its own line, so Rule 1 holds. Every line has at least two points, and A, B, D are in no one line.
- Rule 1. A and C are two different points, and no line contains both.
- 3 lines through each point: each point pairs with each of the other 3, and each pair is its own line.
- Rule 3 gives points A, B, C not all on one line, and Rule 1 gives lines \overleftrightarrow{AB}, \overleftrightarrow{BC} and \overleftrightarrow{AC}. If two were the same line, it would contain A, B and C, which Rule 3 forbids.
- Take any line \ell. Rule 3 gives points A, B, C that do not all lie on one line, so they do not all lie on \ell: at least one of them is a point not on \ell.
- Each defined word leads to another in its definition, so the chain never stops. With finitely many words some word appears twice, and is then defined, through the chain, in terms of itself.
- No. The dot is a model of the point, not the point. The point is the exact location the dot marks; the width belongs to the drawing, not to the geometry.
- Only one line passes through B and C, so the line holding A, B, C and the line holding B, C, D are both that line, and it holds all four. If B = C, the two lines need only share one point, and could be different lines crossing there.
- A line lies in many planes: the plane through two of its points and any point off it contains the whole line, so collinear points are coplanar. Coplanar points need not be collinear: A, B and D all lie in \mathcal{M}, but D is off \ell.