Points, lines, line segments, rays — notation and differences: Worked examples

This is the parent topic's workshop page. The parent taught you the four objects (point, line, ray, segment) and their parametric formulas. Here we do nothing but worked examples — one for every kind of situation the topic can throw at you, so that no exam question ever surprises you.

We will lean on the parametric idea a lot, so let us pin it down in one picture before anything else.

Figure — Points, lines, line segments, rays — notation and differences

Read the figure above: the dashed cyan track is the full line (every $t$ ), the solid cyan portion is the ray (only $t \geq 0$ , forward from $A$ ), and the thick amber piece is the segment (only $0 \leq t \leq 1$ , trapped between $A$ and $B$ ). Notice the labelled dots: $A$ sits at $t = 0$ , $B$ at $t = 1$ , the cyan dot at $t = 2$ lies past $B$ (on ray and line but off the segment), and the cyan dot at $t = - 1$ lies behind $A$ (on the line only). That single picture is the whole page in miniature — every later example is just choosing a cage for $t$ .

Here $B - A$ means "subtract matching coordinates": if $A = (x_{1}, y_{1})$ and $B = (x_{2}, y_{2})$ then $B - A = (x_{2} - x_{1}, y_{2} - y_{1})$ . This little arrow $v$ is called the direction vector — it is literally "how far right, how far up" you travel to get from $A$ to $B$ . (More on arrows in Vectors.)

The scenario matrix

Every question about these objects lands in one of the cells below. The last column tells you which worked example covers it.

Cell	What makes it tricky	Covered by
A. Read the notation	one arrow vs two vs a bar	Ex 1
B. Direction has negative signs	vector pointing left / down	Ex 2
C. Is a point ON the object?	check the slider value $t$	Ex 3
D. Degenerate: $A = B$	zero direction vector	Ex 4
E. Vertical / horizontal	one coordinate never changes	Ex 5
F. Order matters?	$A B$ vs $B A$ vs $\overline{A B}$	Ex 6
G. Real-world word problem	laser / walkway modelling	Ex 7
H. Exam twist / limiting case	segment "grows into" a ray	Ex 8
I. Three points — do they line up?	shared direction vector	Ex 9

Prerequisite links you may want open: Distance Formula, Coordinate Geometry, Colinear Points, Midpoint Formula.

Worked examples

Worked example Example 1 — Cell A: Reading the notation

Statement. Points $P$ , $Q$ , $R$ are marked. Describe in words: (a) $P Q$ , (b) $\overline{QR}$ , (c) $R P$ .

Forecast: Before reading on — which one is infinite in both directions? Which one starts and stops?

(a) $P Q$ = ray, starts at $P$ , shoots through $Q$ , keeps going forever past $Q$ . Why this step? One arrow $=$ one infinite direction; the first letter is the endpoint (the arrow "leaves" $P$ ).
(b) $\overline{QR}$ = segment, the finite piece from $Q$ to $R$ including both dots. Why this step? A bar with no arrows means it stops at both ends — it has a measurable length.
(c) $R P$ = line, straight and endless in both directions through $R$ and $P$ . Why this step? Two arrowheads $=$ infinite both ways, so order and endpoints do not matter.

Verify: Count the arrowheads: ray $1$ , segment $0$ , line $2$ . Matches "one/zero/two infinite ends." ✓

Worked example Example 2 — Cell B: Negative-sign direction vector

Statement. $A = (5, 3)$ , $B = (2, 7)$ . Find the direction vector $v = B - A$ and write the ray $A B$ parametrically.

Forecast: Is $B$ to the left of $A$ or the right? Will the first component of $v$ be positive or negative?

Subtract coordinates: $v = (2 - 5, 7 - 3) = (- 3, 4)$ . Why this step? The direction vector is always "destination minus start." A negative first entry just means we travel left; a positive second entry means we travel up.
Attach to $A$ with a forward slider: $(x, y) = (5, 3) + t (- 3, 4), t \geq 0.$ Why this step? A ray keeps $t \geq 0$ so it only walks in the $v$ direction (up-left here), never backward.

Verify: Put $t = 1$ : $(5, 3) + (- 3, 4) = (2, 7) = B$ . ✓ The point $B$ is reached at $t = 1$ as it must be. Put $t = 2$ : $(5, 3) + (- 6, 8) = (- 1, 11)$ — further up-left, still on the ray. ✓

Read the figure above: the amber arrow is $v = (- 3, 4)$ leaving the endpoint $A = (5, 3)$ and pointing up and to the left — that is what the negative first entry buys us. The cyan ray marches in that same up-left direction: $B$ appears at $t = 1$ and the cyan dot at $t = 2 = (- 1, 11)$ sits even further along, confirming the ray never turns around.

Worked example Example 3 — Cell C: Is a given point on the object?

Statement. Segment $\overline{A B}$ with $A = (1, 2)$ , $B = (7, 10)$ . Is $C = (4, 6)$ on the segment? Is $D = (13, 18)$ on it?

Forecast: $C$ looks like it might be the middle. Will its slider value land inside $0 \leq t \leq 1$ ? What about $D$ ?

Direction vector: $v = B - A = (7 - 1, 10 - 2) = (6, 8)$ . Why this step? We need the slider equation $P (t) = A + t v$ to solve for $t$ .
Solve for $t$ using $C$ 's $x$ -coordinate: $4 = 1 + 6 t \Rightarrow 6 t = 3 \Rightarrow t = \frac{1}{2}$ . Why this step? Each coordinate must give the same $t$ if the point truly lies on the line.
Check the $y$ -coordinate: $2 + 8 \cdot \frac{1}{2} = 2 + 4 = 6$ . ✓ Matches $C_{y} = 6$ . Why this step? If $x$ and $y$ disagree on $t$ , the point is off the line entirely.
Range test for $C$ : $t = \frac{1}{2}$ lies in $0 \leq t \leq 1$ , so $C$ is on the segment (it is the midpoint, in fact). Why this step? Solving for $t$ only tells us the point is on the endless line; the segment is the cage $0 \leq t \leq 1$ , so we must confirm $t$ falls inside that cage before claiming it is on the segment.
Now $D = (13, 18)$ : $13 = 1 + 6 t \Rightarrow 6 t = 12 \Rightarrow t = 2$ ; check $y$ : $2 + 8 \cdot 2 = 18$ . ✓ Same $t$ , so $D$ is on the line — but $t = 2 > 1$ , so $D$ is not on the segment (it lies beyond $B$ ). Why this step? Cell C is exactly about the difference: same algebra, but the range of $t$ decides segment vs ray vs line.

Verify: Midpoint formula gives $(\frac{1 + 7}{2}, \frac{2 + 10}{2}) = (4, 6) = C$ . ✓ And $D$ at $t = 2$ would sit on the ray $A B$ but off the finite segment. ✓

Worked example Example 4 — Cell D: Degenerate case

$A = B$ Statement. Someone writes $A B$ with $A = (3, 3)$ and $B = (3, 3)$ . What object is this?

Forecast: If start and end are the same dot, is there any direction to shoot in?

Direction vector: $v = B - A = (3 - 3, 3 - 3) = (0, 0)$ . Why this step? The zero vector has no direction — there is no "which way" to point.
Plug into the ray formula: $(x, y) = (3, 3) + t (0, 0) = (3, 3)$ for every $t$ . Why this step? Multiplying the zero vector by any $t$ still gives zero, so nothing ever moves.
Conclusion: the "ray" collapses to a single point $(3, 3)$ . Why this step? This is the degenerate limit — when both defining points coincide, all three objects (line, ray, segment) shrink to one point. A genuine ray/line/segment requires two distinct points.

Verify: Length $= 0^{2} + 0^{2} = 0$ . ✓ Zero length confirms it is not a proper segment — see Euclidean Postulates (two distinct points determine exactly one line). ✓

Worked example Example 5 — Cell E: Vertical and horizontal cases

Statement. (a) Line through $A = (4, 1)$ and $B = (4, 9)$ . (b) Segment through $C = (- 2, 5)$ and $D = (6, 5)$ . Describe each and give its length where finite.

Forecast: In (a) does $x$ ever change? In (b) does $y$ ever change?

(a) Direction: $v = (4 - 4, 9 - 1) = (0, 8)$ . The first entry is $0$ , so $x$ is frozen at $4$ . Why this step? A zero horizontal component means a vertical line — every point has $x = 4$ .
(a) Parametric: $(x, y) = (4, 1) + t (0, 8)$ , $t \in R$ , i.e. $x = 4$ , $y = 1 + 8 t$ . As $t$ ranges over all reals, $y$ hits every value — an endless vertical line. Why this step? Writing the parametric form makes the "frozen $x$ " explicit: the $x$ -equation carries no $t$ at all, so no choice of $t$ can ever move us off $x = 4$ , while the $y$ -equation still sweeps the whole axis.
(b) Direction: $v = (6 - (- 2), 5 - 5) = (8, 0)$ . The second entry is $0$ , so $y$ is frozen at $5$ — a horizontal segment. Why this step? Zero vertical component means flat; only $x$ moves.
(b) Length: $∣ \overline{C D} ∣ = (6 - (- 2))^{2} + (5 - 5)^{2} = 8^{2} + 0 = 8$ . Why this step? The Distance Formula still works; the zero term just drops out.

Verify: (a) At $t = 1$ : $(4, 9) = B$ ✓. (b) $64 = 8$ ✓ — matches counting units from $x = - 2$ to $x = 6$ , which is $8$ steps. ✓

Read the figure above: the dashed cyan line runs straight up through $x = 4$ — its $x$ never budges, which is the visual meaning of a "frozen" first coordinate. The thick amber segment lies flat at $y = 5$ ; the double-headed arrow between $C = (- 2, 5)$ and $D = (6, 5)$ marks off the $8$ units of length. One picture, both special cases.

Worked example Example 6 — Cell F: Does the order of letters matter?

Statement. With $A = (0, 0)$ , $B = (6, 2)$ , decide whether each pair is equal: (a) $A B$ vs $B A$ , (b) $A B$ vs $B A$ , (c) $\overline{A B}$ vs $\overline{B A}$ .

Forecast: Which object cares about where you start? Which only cares about which two dots?

(a) Ray: $A B$ starts at $A = (0, 0)$ , direction $(6, 2)$ , going right-up. $B A$ starts at $B = (6, 2)$ , direction $A - B = (- 6, - 2)$ , going left-down. Different endpoints, opposite directions → NOT equal. Why this step? A ray is defined by its endpoint plus direction; both differ here.
(b) Line: both contain all points $A + t (6, 2)$ over all $t \in R$ . Swapping the letters just relabels the same endless track. Equal. Why this step? A line has no start, so naming order is irrelevant.
(c) Segment: $\overline{A B}$ and $\overline{B A}$ both mean "the finite piece with endpoints $A$ and $B$ ." Same two endpoints → same segment. Equal. Why this step? A segment is a set of endpoints, unordered.

Verify: Point $(- 6, - 2)$ (that is $t = - 1$ from $A$ ) is on ray $B A$ but not on ray $A B$ (needs $t = - 1 < 0$ ). So the two rays genuinely differ. ✓ Yet $(- 6, - 2)$ is on the line either way. ✓

Worked example Example 7 — Cell G: Real-world word problem

Statement. A laser is bolted at the point $L = (2, 1)$ (metres, on a floor grid) and aimed so its beam passes through the sensor at $S = (5, 5)$ , continuing until it hits a far wall. (a) Model the beam. (b) A dust speck floats at $(8, 9)$ — is it in the beam? (c) How far is the sensor from the laser?

Forecast: A laser fires forward only — which object (line, ray, segment) matches "starts at a bolt, goes forever one way"?

Choose the object: a ray, endpoint $L$ , because the beam starts at the laser and shoots one way, endless. Why this step? No beam goes behind the laser, so we forbid negative sliders.
Direction vector and parametric form: $v = S - L = (5 - 2, 5 - 1) = (3, 4)$ . Beam: $(x, y) = (2, 1) + t (3, 4)$ , $t \geq 0$ . Why this step? "Destination minus start" gives the way the beam points; attaching it to $L$ with a forward-only slider ( $t \geq 0$ ) captures a beam that leaves the laser and never comes back — exactly a ray.
(b) Test the speck $(8, 9)$ : $8 = 2 + 3 t \Rightarrow t = 2$ ; check $y$ : $1 + 4 \cdot 2 = 9$ . ✓ Same $t = 2 \geq 0$ , so the speck is in the beam (twice as far out as the sensor). Why this step? Both coordinates agree on $t$ , and $t \geq 0$ means it is on the forward ray.
(c) Distance $L$ to $S$ : $(5 - 2)^{2} + (5 - 1)^{2} = 9 + 16 = 25 = 5$ metres. Why this step? The Distance Formula turns the $(3, 4)$ legs into a hypotenuse — a classic $3$ - $4$ - $5$ triangle.

Verify: Speck at $t = 2$ means it is $2 \times 5 = 10$ m from the laser. Direct check: $(8 - 2)^{2} + (9 - 1)^{2} = 36 + 64 = 100 = 10$ m. ✓

Worked example Example 8 — Cell H: Exam twist / limiting case

Statement. Start with segment $\overline{A B}$ , $A = (0, 0)$ , $B = (1, 0)$ . Keep $A$ fixed and slide $B$ to $(2, 0)$ , then $(10, 0)$ , then imagine $B \to (N, 0)$ as $N \to \infty$ . What object does the segment "become," and what happens to its length?

Forecast: If one endpoint runs away forever, do we still have two endpoints?

Direction stays rightward: $v = (N, 0) - (0, 0) = (N, 0)$ , always pointing along the positive $x$ -axis. Why this step? Only the length of $v$ grows; the direction $(1, 0)$ (after scaling) is unchanged.
Measure by real distance instead of the slider. Let $s$ be the ==distance travelled from $A$ == in metres, so a point of the segment is $(s, 0)$ with $0 \leq s \leq N$ (at $s = 0$ we are at $A$ , at $s = N$ we are at $B$ ). Why this step? The slider $t$ always runs $0 \to 1$ no matter how long the segment is, which hides the growth. Re-labelling by the honest physical distance $s$ makes the far endpoint $s = N$ visibly march off to the right, so the limit is easy to see.
Take the limit $N \to \infty$ : the allowed range becomes $0 \leq s < \infty$ — the segment loses its far endpoint and turns into a ray starting at $A$ and pointing along $+ x$ . Why this step? As $N$ grows without bound there is no finite value the far endpoint settles on, so the " $s \leq N$ " cap simply disappears and leaves only the lower stop $s \geq 0$ — that one-sided restriction is precisely the definition of a ray, so the limiting shape must be a ray.
Length: $∣ \overline{A B} ∣ = N^{2} + 0 = N \to \infty$ . Why this step? The measurable length diverges — which is exactly why rays and lines are called "infinite" and have no length.

Verify: For $N = 10$ , length $= 1 0^{2} = 10$ ; for $N = 100$ , length $= 100$ — unbounded as $N$ grows. ✓ The limiting object has one endpoint ( $A = (0, 0)$ ) and no far end: a ray, which we could name $A B^{'}$ using any second point $B^{'} = (1, 0)$ on it. ✓

Worked example Example 9 — Cell I: Do three points line up (collinear)?

Statement. Are $P = (1, 2)$ , $Q = (3, 6)$ , $R = (6, 12)$ collinear (all on one line)? If yes, order them along the line.

Forecast: If they share one straight track, the direction from $P$ to $Q$ should match the direction from $P$ to $R$ . Will the vectors be scalar multiples?

Vector $P \to Q$ : $(3 - 1, 6 - 2) = (2, 4)$ .
Vector $P \to R$ : $(6 - 1, 12 - 2) = (5, 10)$ . Why these steps? Three points are collinear exactly when these two direction vectors are parallel — one is a number times the other (see Parallel and Perpendicular Lines).
Test parallelism (cross-check): $(2) (10) - (4) (5) = 20 - 20 = 0$ . Why this step? This "cross-product" number is zero precisely when the vectors point the same (or opposite) way. Zero ⇒ collinear.
Order them: find $t$ for each along $P + t (2, 4)$ : $P$ at $t = 0$ , $Q$ at $t = 1$ (since $(2, 4) = 1 \cdot (2, 4)$ ), $R$ at $t = 2.5$ (since $(5, 10) = 2.5 \cdot (2, 4)$ ). Increasing $t$ : $P, Q, R$ . Why this step? The slider value orders points along the line just like a ruler.

Verify: $R$ at $t = 2.5$ : $(1, 2) + 2.5 (2, 4) = (1 + 5, 2 + 10) = (6, 12) = R$ . ✓ Cross-check number $= 0$ confirms collinearity. ✓

Recall Quick self-test (reveal after answering)

The slider $t$ for a segment is restricted to which range? ::: $0 \leq t \leq 1$ $A B$ and $B A$ — equal or not? ::: Not equal (opposite endpoints and directions) A point $C$ gives $t = 1.4$ on line $A B$ . On the segment $\overline{A B}$ ? ::: No, because $1.4 > 1$ If $A = B$ , what does $A B$ collapse to? ::: A single point (zero direction vector) Length of the horizontal segment from $(- 2, 5)$ to $(6, 5)$ ? ::: $8$