High School Mathematics · Grades 10–12

15Vectors and Lines in the Plane

Vectors encode displacements; lines are their trajectories. The key new tool of this chapter is the determinant, a single number that decides whether two vectors are collinear — and, from there, produces equations of lines and settles alignment and parallelism questions by pure computation. The same ideas extend to space in Chapter 31.

15.1 Vectors: a recap

Definition 15.1 (Vector, coordinates)

A vector $\vec u$ represents a displacement: all arrows with the same direction, orientation and length represent the same vector. In a coordinate system, $\vec u\,(x, y)$ records the displacement ( $x$ horizontally, $y$ vertically); for points $A(x_A, y_A)$ and $B(x_B, y_B)$ ,

\vect{AB}\,\bigl(x_B - x_A,\ y_B - y_A\bigr).

Vectors add coordinate-wise, and $\lambda\vec u$ has coordinates $(\lambda x, \lambda y)$ .

Proposition 15.2 (Midpoint)

The midpoint $M$ of the segment $[AB]$ has coordinates $\left(\frac{x_A + x_B}{2},\ \frac{y_A + y_B}{2}\right)$ .

Proof. $M$ is characterized by $\vect{AM} = \frac12\vect{AB}$ : coordinate-wise, $x_M - x_A = \frac12(x_B - x_A)$ , so $x_M = \frac{x_A + x_B}{2}$ , and similarly for $y_M$ . ∎

15.2 Collinearity and the determinant

Definition 15.3 (Collinear vectors)

Two vectors are collinear if one is a scalar multiple of the other: $\vec v = \lambda \vec u$ for some $\lambda \in \R$ (or $\vec u = \vec 0$ ). Geometrically: they have the same direction, or one of them is zero.

Definition 15.4 (Determinant)

The determinant of $\vec u\,(x, y)$ and $\vec v\,(x', y')$ is

\det(\vec u, \vec v) = \begin{vmatrix} x & x' \\ y & y' \end{vmatrix} = x y' - x' y .

Theorem 15.5 (Collinearity criterion)

Two vectors $\vec u$ and $\vec v$ are collinear if and only if $\det(\vec u, \vec v) = 0$ .

Proof. ( $\Rightarrow$ ) If $\vec v = \lambda\vec u$ , then $x' = \lambda x$ and $y' = \lambda y$ , so $\det(\vec u, \vec v) = x(\lambda y) - (\lambda x)y = 0$ ; if $\vec u = \vec 0$ , the determinant is trivially $0$ .

( $\Leftarrow$ ) Suppose $xy' - x'y = 0$ and $\vec u \neq \vec 0$ , say $x \neq 0$ (the case $y \neq 0$ is symmetric). Set $\lambda = \frac{x'}{x}$ . Then $x' = \lambda x$ by construction, and the relation $xy' = x'y = \lambda x y$ gives, after dividing by $x \neq 0$ , $y' = \lambda y$ . Hence $\vec v = \lambda \vec u$ . ∎

Example 15.6

$\vec u\,(3, -2)$ and $\vec v\,(-6, 4)$ : $\det = 3 \times 4 - (-6) \times (-2) = 12 - 12 = 0$ , collinear (indeed $\vec v = -2\vec u$ ). $\vec u\,(3, -2)$ and $\vec w\,(2, 1)$ : $\det = 3 + 4 = 7 \neq 0$ , not collinear.

Method 15.7 (Alignment of three points)

$A$ , $B$ , $C$ are aligned if and only if $\vect{AB}$ and $\vect{AC}$ are collinear: compute both vectors and check $\det\bigl(\vect{AB}, \vect{AC}\bigr) = 0$ .

15.3 Cartesian equations of lines

Definition 15.8 (Direction vector)

A direction vector of a line $d$ is any nonzero vector $\vec u$ such that $\vect{AB}$ is collinear with $\vec u$ for all points $A, B$ of $d$ .

Theorem 15.9 (Cartesian equation)

A line with direction vector $\vec u\,(-b, a)$ admits an equation of the form

ax + by + c = 0 \qquad (a, b) \neq (0, 0).

Conversely, every such equation describes a line with direction vector $\vec u\,(-b, a)$ .

Proof. Let $d$ pass through $A(x_A, y_A)$ with direction $\vec u\,(-b, a)$ . A point $M(x, y)$ lies on $d$ exactly when $\vect{AM}$ is collinear with $\vec u$ , i.e. (Theorem 15.5)

0 = \det\bigl(\vect{AM}, \vec u\bigr) = (x - x_A)\,a - (-b)(y - y_A) = ax + by - (a x_A + b y_A),

an equation of the announced form with $c = -(a x_A + b y_A)$ . Conversely, the solutions of $ax + by + c = 0$ with, say, $b \neq 0$ are the points $\left(x, -\frac{a x + c}{b}\right)$ : subtracting two of them shows every chord is collinear with $(-b, a)$ , and one solution always exists — it is the line through it directed by $(-b, a)$ . ∎

Method 15.10 (Line through two points)

To find an equation of the line $(AB)$ : compute the direction vector $\vect{AB}\,(x_B - x_A, y_B - y_A)$ ; match it with $(-b, a)$ , i.e. take $a = y_B - y_A$ and $b = -(x_B - x_A)$ ; determine $c$ by plugging in the coordinates of $A$ .

Example 15.11

Line through $A(1, 2)$ and $B(4, 3)$ : $\vect{AB}\,(3, 1)$ , so $a = 1$ , $b = -3$ , and the equation is $x - 3y + c = 0$ . Plugging in $A$ : $1 - 6 + c = 0$ , so $c = 5$ . Equation: $x - 3y + 5 = 0$ . Check with $B$ : $4 - 9 + 5 = 0$ .

Remark 15.12

When $b \neq 0$ , the equation can be solved for $y$ : $y = mx + p$ with slope $m = -\frac ab$ . The cartesian form $ax + by + c = 0$ is more general: it also covers vertical lines ( $b = 0$ ), which have no slope.

The line x - 3y + 5 = 0 of , its direction vector u = AB (red, here scaled by 1/2), and the two points used to build the equation. — The line $x - 3y + 5 = 0$ of Example 15.11, its direction vector $\vec u = \vect{AB}$ (red, here scaled by $\frac12$ ), and the two points used to build the equation.

15.4 Parametric representation

Definition 15.13 (Parametric representation)

The line through $A(x_A, y_A)$ with direction $\vec u\,(\alpha, \beta)$ is the set of points

M\bigl(x_A + t\alpha,\ y_A + t\beta\bigr), \qquad t \in \R .

The number $t$ is the parameter: each value of $t$ produces one point of the line, as if travelling along it at constant speed $\vec u$ .

Example 15.14

The line of Example 15.11 is also $\{(1 + 3t,\ 2 + t) : t \in \R\}$ : $t = 0$ gives $A$ , $t = 1$ gives $B$ . Eliminating $t$ ( $t = y - 2$ , so $x = 1 + 3(y-2)$ ) recovers $x - 3y + 5 = 0$ .

15.5 Relative positions of two lines

Proposition 15.15 (Parallel or intersecting)

Two lines with direction vectors $\vec u$ and $\vec v$ are parallel if and only if $\det(\vec u, \vec v) = 0$ . In equation form: $d : ax + by + c = 0$ and $d' : a'x + b'y + c' = 0$ are parallel if and only if $ab' - a'b = 0$ . Non-parallel lines meet in exactly one point.

Proof. Parallelism means collinear directions, which is Theorem 15.5; with $\vec u\,(-b, a)$ and $\vec v\,(-b', a')$ , the determinant is $(-b)a' - (-b')a = ab' - a'b$ . If the lines are not parallel, solving the two equations simultaneously (by substitution or combination) leads to a unique solution: the system behaves like a nonzero-determinant $2 \times 2$ system. ∎

Method 15.16 (Intersection of two lines)

Solve the system of the two equations: isolate one unknown in one equation and substitute into the other, or combine the equations to eliminate one unknown. Always check the resulting point in both equations.

Example 15.17

Intersect $d: x - 3y + 5 = 0$ and $d': 2x + y - 4 = 0$ . From $d'$ : $y = 4 - 2x$ . Substituting into $d$ :

x - 3(4 - 2x) + 5 = 0 \iff x - 12 + 6x + 5 = 0 \iff 7x = 7 \iff x = 1,

then $y = 2$ . The lines meet at $(1, 2)$ . Check in $d$ : $1 - 6 + 5 = 0$ .

15.6 Exercises

Exercise 15.1 ★

Given $A(2, -1)$ , $B(5, 3)$ and $C(-1, 1)$ , compute the coordinates of $\vect{AB}$ , $\vect{AC}$ , $\vect{AB} + \vect{AC}$ and $2\vect{AB} - 3\vect{AC}$ , and the midpoint of $[BC]$ .

Solution

Solution of Exercise 15.1.

$\vect{AB}\,(3, 4)$ , $\vect{AC}\,(-3, 2)$ , $\vect{AB} + \vect{AC}\,(0, 6)$ , $2\vect{AB} - 3\vect{AC}\,(6 + 9,\ 8 - 6) = (15, 2)$ . The midpoint of $[BC]$ is $\left(\frac{5 + (-1)}{2}, \frac{3+1}{2}\right) = (2, 2)$ .

Exercise 15.2 ★

Are the vectors collinear?

\vec u\,(4, -6) \text{ and } \vec v\,(-2, 3); \qquad \vec u\,(2, 5) \text{ and } \vec v\,(3, 7); \qquad \vec u\,(0, 3) \text{ and } \vec v\,(0, -8).

Solution

Solution of Exercise 15.2.

$\det = 4 \times 3 - (-2) \times (-6) = 12 - 12 = 0$ : collinear (indeed $\vec v = -\frac12 \vec u$ ).

$\det = 2 \times 7 - 3 \times 5 = -1 \neq 0$ : not collinear.

$\det = 0 \times (-8) - 0 \times 3 = 0$ : collinear (both vertical).

Exercise 15.3 ★

Find a cartesian equation of the line through $A(2, 1)$ with direction vector $\vec u\,(3, -1)$ , then of the line through $B(0, 4)$ and $C(2, 0)$ .

Solution

Solution of Exercise 15.3.

Through $A(2,1)$ , direction $(3, -1)$ : matching $(-b, a) = (3, -1)$ gives $a = -1$ , $b = -3$ , so $-x - 3y + c = 0$ ; plugging in $A$ : $-2 - 3 + c = 0$ , $c = 5$ . Multiplying by $-1$ : $x + 3y - 5 = 0$ .

Through $B(0,4)$ and $C(2,0)$ : $\vect{BC}\,(2, -4)$ , so $a = -4$ , $b = -2$ : $-4x - 2y + c = 0$ ; plugging in $B$ : $-8 + c = 0$ , $c = 8$ . Simplifying by $-2$ : $2x + y - 4 = 0$ . Check with $C$ : $4 + 0 - 4 = 0$ .

Exercise 15.4 ★

For the line $d: 3x - 2y + 6 = 0$ , give a direction vector, the intersections with the two coordinate axes, and decide whether the points $P(2, 6)$ and $Q(1, 4)$ belong to $d$ .

Solution

Solution of Exercise 15.4.

A direction vector is $(-b, a) = (2, 3)$ . Intersections: $x = 0$ gives $y = 3$ , point $(0, 3)$ ; $y = 0$ gives $x = -2$ , point $(-2, 0)$ . For $P(2,6)$ : $3 \times 2 - 2 \times 6 + 6 = 0$ , so $P \in d$ . For $Q(1,4)$ : $3 - 8 + 6 = 1 \neq 0$ , so $Q \notin d$ .

Exercise 15.5 ★★

Are the points $A(1, 2)$ , $B(4, 8)$ , $C(-2, -4)$ aligned? Same question for $D(0, 1)$ , $E(2, 4)$ , $F(5, 9)$ .

Solution

Solution of Exercise 15.5.

$\vect{AB}\,(3, 6)$ and $\vect{AC}\,(-3, -6)$ : $\det = 3 \times (-6) - (-3) \times 6 = 0$ , so $A$ , $B$ , $C$ are aligned.

$\vect{DE}\,(2, 3)$ and $\vect{DF}\,(5, 8)$ : $\det = 2 \times 8 - 5 \times 3 = 1 \neq 0$ : $D$ , $E$ , $F$ are not aligned.

Exercise 15.6 ★★

Determine the intersection point, if any:

d: 2x + y - 7 = 0 \ \text{ and }\ d': x - y + 1 = 0; \qquad e: x - 2y + 3 = 0 \ \text{ and }\ e': -2x + 4y + 1 = 0 .

Solution

Solution of Exercise 15.6.

First pair: adding the two equations, $(2x + y - 7) + (x - y + 1) = 3x - 6 = 0$ , so $x = 2$ , then $y = x + 1 = 3$ : they meet at $(2, 3)$ (check: $4 + 3 - 7 = 0$ ).

Second pair: $ab' - a'b = 1 \times 4 - (-2) \times (-2) = 0$ , so the lines are parallel; multiplying $e$ by $-2$ gives $-2x + 4y - 6 = 0$ , which differs from $e'$ ( $-6 \neq 1$ ): strictly parallel, no intersection.

Exercise 15.7 ★★

Give a parametric representation of the line $d: 2x - y + 3 = 0$ , and a cartesian equation of the line $\bigl\{(1 + 2t,\ -3 + t) : t \in \R\bigr\}$ .

Solution

Solution of Exercise 15.7.

For $d: 2x - y + 3 = 0$ : direction $(-b, a) = (1, 2)$ and the point $(0, 3)$ lies on $d$ , so $d = \{(t,\ 3 + 2t) : t \in \R\}$ .

For the parametric line through $(1, -3)$ with direction $(2, 1)$ : $a = 1$ , $b = -2$ , so $x - 2y + c = 0$ ; plugging in $(1, -3)$ : $1 + 6 + c = 0$ , $c = -7$ . Equation: $x - 2y - 7 = 0$ (check $t = 1$ , point $(3, -2)$ : $3 + 4 - 7 = 0$ ).

Exercise 15.8 ★★

Find the equation of the line through $P(1, -2)$ parallel to $d: 4x - 3y + 2 = 0$ , and the value of $m$ for which the line $mx + 2y - 5 = 0$ is parallel to $d$ .

Solution

Solution of Exercise 15.8.

A parallel line has the same $(a, b)$ : $4x - 3y + c = 0$ ; through $P(1, -2)$ : $4 + 6 + c = 0$ , so $c = -10$ and the line is $4x - 3y - 10 = 0$ .

Parallelism of $mx + 2y - 5 = 0$ with $d$ requires $ab' - a'b = 4 \times 2 - m \times (-3) = 8 + 3m = 0$ , so $m = -\frac83$ .

Exercise 15.9 ★★

Let $A(-1, 0)$ , $B(3, 2)$ and $C(1, -4)$ . Find a cartesian equation of the median of the triangle $ABC$ issued from $C$ (the line joining $C$ to the midpoint of $[AB]$ ).

Solution

Solution of Exercise 15.9.

The midpoint of $[AB]$ is $M\!\left(\frac{-1+3}{2}, \frac{0+2}{2}\right) = (1, 1)$ . The median is the line $(CM)$ with $C(1, -4)$ : both points have abscissa $1$ , so the median is the vertical line

x - 1 = 0 .

Exercise 15.10 ★★

For which value(s) of the parameter $m$ are the vectors $\vec u\,(m, 2)$ and $\vec v\,(3, m - 1)$ collinear?

Solution

Solution of Exercise 15.10.

$\det(\vec u, \vec v) = m(m - 1) - 3 \times 2 = m^2 - m - 6$ . It vanishes when $m^2 - m - 6 = 0$ : $\Delta = 1 + 24 = 25$ , roots $m = \frac{1 \pm 5}{2}$ , i.e. $m = 3$ or $m = -2$ .

Exercise 15.11 ★★★

Let $ABCD$ be a parallelogram, $I$ the midpoint of $[AB]$ , and $J$ the point defined by $\vect{DJ} = \frac23 \vect{DI}$ . Working in the coordinate system where $A(0,0)$ , $B(1,0)$ , $D(0,1)$ (so $C(1,1)$ ):

Compute the coordinates of $I$ and $J$ .
Show that $A$ , $J$ and $C$ are aligned. (A nice fact: the line $(DI)$ cuts the diagonal $(AC)$ one third of the way up.)

Solution

Solution of Exercise 15.11.

1. $I = \left(\frac12, 0\right)$ . Then $\vect{DI} = \left(\frac12 - 0,\ 0 - 1\right) = \left(\frac12, -1\right)$ and $\vect{DJ} = \frac23\vect{DI} = \left(\frac13, -\frac23\right)$ , so

J = D + \vect{DJ} = \left(\frac13,\ \frac13\right).

2. $\vect{AJ}\left(\frac13, \frac13\right)$ and $\vect{AC}\,(1, 1)$ : $\det = \frac13 \times 1 - 1 \times \frac13 = 0$ , so $A$ , $J$ , $C$ are aligned — in fact $\vect{AJ} = \frac13\vect{AC}$ : the line $(DI)$ cuts the diagonal $(AC)$ at exactly one third of its length from $A$ .

15.7 Problem: Collision courses and clever coordinates

Problem 15.1

Weekend problem — crossing paths is not colliding: closest approach at sea, and theorems proved by choosing the right frame

Two ships’ straight courses cross — must the ships collide? Not unless they reach the crossing at the same time: paths are sets of points, motions are parametric, and confusing the two has sunk real ships. This problem runs a small vessel-traffic service on parametric lines (Definition 15.13), then turns to pure geometry with the other superpower of coordinates: choosing them cleverly, as in Exercise 15.11, to make classical theorems compute themselves.

Part I — Lines, three ways.

Give a parametric representation of the line through $A(1, 2)$ with direction vector $\vec u(3, -1)$ , then eliminate the parameter to obtain a Cartesian equation.
For the line $2x - 5y + 3 = 0$ : read off a direction vector (Theorem 15.9), find one point, and write a parametric representation.
Compute the intersection point of the two lines of questions 1 and 2 (Method 15.16).
Confirm with the determinant (Theorem 15.5) that those two lines had to intersect; then show that $2x - 5y + 3 = 0$ and $-4x + 10y + 1 = 0$ are strictly parallel.
Are $A(2, 1)$ , $B(5, 3)$ , $C(11, 7)$ aligned (Method 15.7)?

Part II — The vessel-traffic service. At time $t$ (hours), ship $P$ is at $(2t,\ 3 + t)$ and ship $Q$ at $(10 - t,\ 2t - 1)$ (distances in nautical miles).

Find the Cartesian equations of the two paths, and compute where the paths cross.
At what time does each ship pass the crossing point? Do the ships collide? State the general warning in one sentence: crossing paths against meeting motions.
Compute the squared distance $D^2(t)$ between the ships at time $t$ , and reduce it to a quadratic in $t$ . At what time are the ships closest, and how close do they come (Problem 10.1 vertex technology)?
Regulations demand a separation of at least $1$ mile. Is it violated? If so, solve $D^2(t) < 1$ and give the duration of the violation.
Explain why, for any two ships on straight courses at constant speeds, $D^2(t)$ is always a quadratic function of $t$ — so “closest point of approach” is always a vertex computation.
Interception: a patrol boat starts at the origin at $t = 0$ with constant velocity $(a, b)$ and must meet ship $P$ exactly at $t = 3$ . Compute the required $(a, b)$ and the patrol boat’s speed.

Part III — Clever coordinates. Work in the frame of Exercise 15.11: the parallelogram $ABCD$ becomes $A(0,0)$ , $B(1,0)$ , $C(1,1)$ , $D(0,1)$ ; let $I$ be the midpoint of $[AB]$ and $K$ the midpoint of $[CD]$ .

Compute a Cartesian equation of the line $(DI)$ .
Intersect $(DI)$ with the diagonal $(AC)$ : recover Exercise 15.11’s fact — the cut lies exactly one third of the way up the diagonal.
Now the line $(BK)$ : find its equation, intersect with $(AC)$ , and conclude the full classical theorem: the cevians $(DI)$ and $(BK)$ trisect the diagonal $(AC)$ .
Why was it legitimate to prove the theorem in this one special frame? (What does the choice of coordinates preserve, and what does the statement involve?) Answer in two or three sentences.
One more trisection for practice: let $E$ be the midpoint of $[BC]$ . Where does the line $(AE)$ cut the other diagonal $(DB)$ ?

Part IV — Position puzzles.

Are the three lines $x + y = 4$ , $2x - y = 2$ , $x - 2y = -2$ concurrent? (Intersect two, test the third.)
For each real $m$ , let $d_m$ be the line $mx - y + 2 - 3m = 0$ . Show that every $d_m$ passes through one fixed point. (Group the terms in $m$ .)
In the family $(d_m)$ : which member is parallel to $y = 2x + 7$ ? Which passes through the origin?
Finale — the three costumes of a line: Cartesian (membership tests in one substitution), reduced (slope and graph at sight), parametric (motion, timing, interception); the determinant as parallelism oracle; and the well-chosen frame as a theorem factory. One sentence each.

Solution

Solution of Problem 15.1.

1. $x = 1 + 3t$ , $y = 2 - t$ . From the second, $t = 2 - y$ ; substituting: $x = 1 + 3(2 - y) = 7 - 3y$ : Cartesian equation $x + 3y - 7 = 0$ .

2. For $ax + by + c = 0$ , a direction vector is $(-b, a)$ : here $(5, 2)$ . A point: $y = 1$ gives $x = 1$ : $(1, 1)$ . Parametric: $x = 1 + 5t$ , $y = 1 + 2t$ .

3. From $x = 7 - 3y$ in $2x - 5y + 3 = 0$ : $14 - 6y - 5y + 3 = 0$ , so $y = \frac{17}{11}$ and $x = 7 - \frac{51}{11} = \frac{26}{11}$ : the point $\left(\frac{26}{11}, \frac{17}{11}\right)$ .

4. Directions $(5, 2)$ and $(3, -1)$ : $\det = 5 \times (-1) - 2 \times 3 = -11 \neq 0$ : intersecting, as found. For the second pair: directions $(5, 2)$ and $(10, 4)$ have $\det = 0$ (parallel), and doubling $2x - 5y + 3 = 0$ gives $-4x + 10y - 6 = 0 \neq -4x + 10y + 1 = 0$ : strictly parallel.

5. $\vect{AB}(3, 2)$ , $\vect{AC}(9, 6)$ : $\det = 18 - 18 = 0$ : aligned.

6. $P$ : from $x = 2t$ , $y = 3 + t$ : $y = 3 + \frac x2$ . $Q$ : from $x = 10 - t$ : $t = 10 - x$ , so $y = 2(10 - x) - 1 = 19 - 2x$ . Crossing: $3 + \frac x2 = 19 - 2x$ gives $x = 6.4$ , $y = 6.2$ : the paths cross at $(6.4,\ 6.2)$ .

7. $P$ passes there when $2t = 6.4$ : $t = 3.2$ h. $Q$ when $10 - t = 6.4$ : $t = 3.6$ h. Twenty-four minutes apart: no collision. Warning: a chart shows paths; only the parametric clocks say whether two motions occupy the same point at the same instant.

8. $D^2(t) = (3t - 10)^2 + (4 - t)^2 = 10t^2 - 68t + 116$ : vertex at $t = \frac{68}{20} = 3.4$ h, where $D^2 = 0.4$ : closest approach $D = \sqrt{0.4} \approx 0.63$ mile at $t = 3.4$ .

9. Violated: $0.63 < 1$ . $D^2(t) < 1$ reads $10t^2 - 68t + 115 < 0$ : $\Delta = 4624 - 4600 = 24$ , roots $\frac{68 \pm \sqrt{24}}{20} = 3.4 \pm 0.245$ : the ships are too close from $t \approx 3.16$ to $t \approx 3.64$ — about $29$ minutes of infraction. One of them must alter course.

10. Each coordinate of each ship is an affine function of $t$ , so each difference of coordinates is affine, and $D^2$ — a sum of two squared affine functions — is a quadratic (with nonnegative leading coefficient). Closest approach is therefore always read off a vertex.

11. Meeting $P$ at $t = 3$ , i.e. at $(6, 6)$ : $(3a, 3b) = (6, 6)$ gives $(a, b) = (2, 2)$ , speed $\sqrt{8} \approx 2.8$ knots.

12. $D(0, 1)$ , $I\left(\frac12, 0\right)$ : direction $\left(\frac12, -1\right)$ , i.e. $(1, -2)$ : equation $2x + y = 1$ .

13. $(AC)$ : $y = x$ . Then $2x + x = 1$ : $x = \frac13$ : the point $\left(\frac13, \frac13\right)$ , one third of the way from $A$ to $C$ .

14. $B(1, 0)$ , $K\left(\frac12, 1\right)$ : direction $\left(-\frac12, 1\right)$ , equation $2x + y = 2$ . With $y = x$ : $x = \frac23$ : the point $\left(\frac23, \frac23\right)$ . The two cevians cut $(AC)$ at its two trisection points: theorem proved, twice three lines of arithmetic.

15. Any parallelogram is carried onto the unit square by choosing $A$ as origin and $\vect{AB}$ , $\vect{AD}$ as the two axes’ units. That choice preserves everything the theorem speaks of — midpoints, alignment, and ratios along a given line (all expressible with vectors and scalars) — so proving the statement in the square proves it for every parallelogram. (It would not transport lengths or angles: those the frame distorts.)

16. $E\left(1, \frac12\right)$ : line $(AE)$ : $y = \frac x2$ . Diagonal $(DB)$ : from $(0,1)$ to $(1,0)$ : $y = 1 - x$ . Intersection: $\frac x2 = 1 - x$ : $x = \frac23$ : the point $\left(\frac23, \frac13\right)$ — again a trisection point, now of the other diagonal. Midpoint cevians love thirds.

17. First two: $x + y = 4$ and $2x - y = 2$ give $3x = 6$ : $(2, 2)$ . Third: $2 - 4 = -2$ : satisfied. Concurrent at $(2, 2)$ .

18. $mx - y + 2 - 3m = m(x - 3) + (2 - y) = 0$ : for the equation to hold for every $m$ , take $x = 3$ , $y = 2$ — and indeed every $d_m$ contains $(3, 2)$ . A pencil of lines through one point, one line per slope $m$ .

19. Parallel to $y = 2x + 7$ : slope $m = 2$ , the line $2x - y - 4 = 0$ . Through the origin: $2 - 3m = 0$ : $m = \frac23$ .

20. Cartesian: plug in a point, membership answered. Reduced: slope and intercept visible, graph immediate. Parametric: the clock built in — collisions, closest approach, interception. The determinant: one multiplication cross, and parallelism is decided. The chosen frame: a parallelogram becomes a unit square, and a classical trisection theorem becomes three intersections of lines.