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

Count the arrowheads — each arrowhead means "infinite in that direction."

• (a) $\overline{PQ}$ — a bar, no arrows → line segment from $P$ to $Q$ (two endpoints, finite).
• (b) $\overrightarrow{PQ}$ — one arrow → ray starting at $P$, through $Q$, forever.
• (c) $\overleftrightarrow{PQ}$ — two arrows → line through $P$ and $Q$ (no endpoints, infinite both ways).

• "A point has length" is False — a point is a location with no dimension (no length, width, or height).
• A ray has exactly one endpoint (the point it starts at). The second letter only shows direction.

Step 1 — direction. Subtract start from end so the arrow points from $A$ toward $B$: $\vec{v}=B-A=(4-1,6-2)=(3,4).$ Why $B-A$? Coordinates measure position from the origin. The journey from $A$ to $B$ is "how much you must add to $A$ to arrive at $B$." Solving $A+\vec{v}=B$ gives $\vec{v}=B-A$. So the $x$-part $(4-1=3)$ is "how far right you travel" and the $y$-part $(6-2=4)$ is "how far up" — together they are the direction from $A$ toward $B$. Step 2 — ray. Start at $A$ (that's $t=0$) and slide forward only, so $t\ge 0$: $(x,y)=(1,2)+t(3,4),t\ge 0.$ Check the endpoints of the parameter: at $t=0$, $(x,y)=(1,2)=A$ ✓; at $t=1$, $(x,y)=(1+3,2+4)=(4,6)=B$ ✓. So $t=0$ is $A$, $t=1$ is $B$, and $t\ge 0$ keeps everything from $A$ onward — that restriction is what makes it a ray and not the whole line.

The two components $(3,4)$ are the legs of a right triangle; the segment is the hypotenuse, so we use the Distance Formula (Pythagoras in coordinates): $|\overline{AB}|=\sqrt{(4-1{)}^2+(6-2{)}^2}=\sqrt{9+16}=\sqrt{25}=5.$ Length $=5$ units.

The segment is $A+t(B-A)$ with $0\le t\le 1$, where $t=0$ gives $A$ and $t=1$ gives $B$ (check: $A+1(B-A)=B$). So $t=\frac{1}{2}$ is exactly halfway between them — the Midpoint Formula: $M=(1,2)+\frac{1}{2}(3,4)=\left(1+1.5,2+2\right)=(2.5,4).$ Check: $2.5=\frac{1+4}{2}$ and $4=\frac{2+6}{2}$. ✓

Step 1 — solve for $t$. We need $C=A+t\vec{v}$, i.e. $(7,10)=(1,2)+t(3,4)$.

• $x$: $7=1+3t\Rightarrow t=2$.
• $y$: $10=2+4t\Rightarrow t=2$. Same $t$ → $C$ is on the line. Step 2 — apply the range. Recall $t=0$ is $A$ and $t=1$ is $B$.
• Ray needs $t\ge 0$: $t=2\ge 0$ ✓ → $C$ is on $\overrightarrow{AB}$ (it's past $B$, which sits at $t=1$).
• Segment needs $0\le t\le 1$: $t=2>1$ ✗ → $C$ is not on $\overline{AB}$. This is exactly why a ray and a segment differ: same line, different allowed $t$.

• Rays differ: $\overrightarrow{AB}$ has endpoint $A$ and goes toward $B$; $\overrightarrow{BA}$ has endpoint $B$ and goes toward $A$. Different starting points, opposite directions → different rays.
• Segments equal: $\overline{AB}$ and $\overline{BA}$ name the same two endpoints; order doesn't matter for a finite piece with both ends fixed → equal.
• Lines equal: a line has no endpoints, so swapping the two naming points changes nothing → equal. See the three rows in the figure below.

Collinearity means "on one straight line." Direction $\overrightarrow{PR}=R-P=(4,6)$. Test if $Q-P=(2,3)$ is a scalar multiple: $(2,3)=t(4,6)\Rightarrow t=\frac{2}{4}=\frac{1}{2},\frac{3}{6}=\frac{1}{2}.$ Both give $t=\frac{1}{2}$ → yes, collinear (Colinear Points). Since $0\le \frac{1}{2}\le 1$, $Q$ lies on segment $\overline{PR}$ (it is in fact the midpoint).

Step 1 — length of $\vec{v}$. $|\vec{v}|=\sqrt{3^2+(-4{)}^2}=\sqrt{9+16}=5$. Step 2 — how many steps of $\vec{v}$ give distance $10$? Each unit of $t$ moves us $|\vec{v}|=5$. To travel $10$ we need $t=\frac{10}{5}=2$ (positive, so it stays on the ray). Step 3 — plug in: $D=(2,1)+2(3,-4)=(2+6,1-8)=(8,-7).$ Check: $|D-A|=\sqrt{6^2+(-8{)}^2}=\sqrt{36+64}=\sqrt{100}=10$ ✓.

Two rays with a common endpoint form an angle. Ray 1 points along the positive $x$-axis, ray 2 along the positive $y$-axis. These directions are perpendicular, so the angle is $90^{\circ }.$ (Their direction vectors $(1,0)$ and $(0,1)$ have dot product $1\cdot 0+0\cdot 1=0$, the hallmark of a right angle.)

Line allows any real $t$. At $t=-1$: $P=(1,2)+(-1)(3,4)=(1-3,2-4)=(-2,-2).$ Since $t=-1<0$, this point is on the opposite side of $A$ from $B$ — it lies on the line but on neither the segment ($0\le t\le 1$) nor the ray $\overrightarrow{AB}$ ($t\ge 0$). It is on the opposite ray $\overrightarrow{A{B}^{\prime }}$ going backwards.

Step 1 — where does $x=10$? From $x=1+3t=10$ we get $t=3$. Step 2 — is that on the ray? $t=3\ge 0$ ✓, so yes, the ray reaches the wall. Step 3 — the point. $y=2+4(3)=14$, so it hits at $(10,14)$. Step 4 — distance. $t=3$ copies of $\vec{v}$ (length $5$) → distance $=3\times 5=15$. Confirm: $\sqrt{(10-1{)}^2+(14-2{)}^2}=\sqrt{81+144}=\sqrt{225}=15$ ✓.

Step 1 — parametrise both.

• $\overline{PQ}$: $(x,y)=(4s,4s)$, $0\le s\le 1$.
• $\overline{RS}$: $(x,y)=(4u,4-4u)$, $0\le u\le 1$. Step 2 — equate. $4s=4u$ and $4s=4-4u$. From the first $s=u$; substitute: $4s=4-4s\Rightarrow 8s=4\Rightarrow s=\frac{1}{2}$ (and $u=\frac{1}{2}$). Step 3 — both in $[0,1]$? $s=u=\frac{1}{2}$ ✓, so the segments do intersect at $(4\cdot \frac{1}{2},4\cdot \frac{1}{2})=(2,2).$ That's the shared centre of the two crossing diagonals.

Length: $|\overline{AB}|=\sqrt{(2-2{)}^2+(9-3{)}^2}=\sqrt{0+36}=6$. Midpoint: $\left(\frac{2+2}{2},\frac{3+9}{2}\right)=(2,6)$. Orientation: the $x$-coordinate is the same ($2=2$) while $y$ changes, so the segment is vertical — every point on it has $x=2$. Direction vector $(0,6)$ points straight up, confirming vertical (see Coordinate Geometry).

Which range of $t$ defines a segment, and which Q used it? ::: $0\le t\le 1$ — e.g. Q5 ($t=\frac{1}{2}$ midpoint) and Q8 ($Q$ on $\overline{PR}$) Which range of $t$ defines a ray from the start point, and which Q pushed past $t=1$? ::: $t\ge 0$ — Q6 had $t=2>1$ (on the ray, off the segment) and Q9 used $t=2$ Which Q produced a point with $t<0$ (off both ray and segment)? ::: Q11, where $t=-1$ landed on the opposite side of $A$ Which range of $t$ defines a whole line? ::: $t\in \mathbb{R}$ (Q11 lives here, since $t=-1$ is still allowed) How do you turn a desired distance $d$ into a parameter along $\vec{v}$? ::: $t=d/|\vec{v}|$ — as in Q9 ($t=10/5=2$) and Q12 ($t=3$ gave distance $15$) Is $\overrightarrow{AB}=\overrightarrow{BA}$? ::: No — opposite endpoints and directions (Q7)