---
title: "Vectors and Lines in the Plane"
book: "High School Mathematics"
subject: math
language: en
chapter: 15
exercises: 11
source: https://one-course.com/books/math/2/en/chapter/15-vectors-and-lines-in-the-plane
---

# Chapter 15 — Vectors and Lines in the Plane

[Vectors](#def-g11-vect-vector) encode displacements; lines are their trajectories. The key new tool of this chapter is the *[determinant](#def-g11-vect-det)*, a single number that decides whether two [vectors](#def-g11-vect-vector) are [collinear](#def-g11-vect-collinear) — and, from there, produces [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of lines and settles alignment and parallelism questions by pure computation. The same ideas extend to space in [Chapter 31](https://one-course.com/books/math/2/en/chapter/31-vectors-lines-and-planes-in-space#ch-g12-space).

## 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $(\lambda x, \lambda y)$.

**Proposition 15.2 (Midpoint).**

The [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) $M$ of the segment $[AB]$ has [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $\left(\frac{x_A + x_B}{2},\ \frac{y_A + y_B}{2}\right)$.

**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](#def-g11-vect-vector) 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](#def-g11-vect-vector) $\vec u$ and $\vec v$ are [collinear](#def-g11-vect-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](#def-g11-vect-det) 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](#def-g11-vect-collinear) (indeed $\vec v = -2\vec u$). $\vec u\,(3, -2)$ and $\vec w\,(2, 1)$: $\det = 3 + 4 = 7 \neq 0$, not [collinear](#def-g11-vect-collinear).

**Method 15.7 (Alignment of three points).**

$A$, $B$, $C$ are aligned if and only if $\vect{AB}$ and $\vect{AC}$ are [collinear](#def-g11-vect-collinear): compute both [vectors](#def-g11-vect-vector) 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](#def-g11-vect-vector) $\vec u$ such that $\vect{AB}$ is [collinear](#def-g11-vect-collinear) with $\vec u$ for all points $A, B$ of $d$.

**Theorem 15.9 (Cartesian equation).**

A line with [direction vector](#def-g11-vect-direction) $\vec u\,(-b, a)$ admits an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the form

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

Conversely, every such [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) describes a line with [direction vector](#def-g11-vect-direction) $\vec u\,(-b, a)$.

**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](#def-g11-vect-collinear) with $\vec u$, i.e. ([Theorem 15.5](#thm-g11-vect-collinearity))

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

an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](#def-g11-vect-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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the line $(AB)$: compute the [direction vector](#def-g11-vect-direction) $\vect{AB}\,(x_B - x_A, y_B - y_A)$; 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) is $x - 3y + c = 0$. Plugging in $A$: $1 - 6 + c = 0$, so $c = 5$. [Equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation): $x - 3y + 5 = 0$. Check with $B$: $4 - 9 + 5 = 0$.

**Remark 15.12.**

When $b \neq 0$, the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) can be solved for $y$: $y = mx + p$ with [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $m = -\frac ab$. The cartesian form $ax + by + c = 0$ is more general: it also covers vertical lines ($b = 0$), which have no [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine).

![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.](https://one-course.com/images/onecourse/chapters/math-2/g11-vect/fig-0c3d7a0cf8c8.svg)

*The line $x - 3y + 5 = 0$ of [Example 15.11](#ex-g11-vect-twopoints), its [direction vector](#def-g11-vect-direction) $\vec u = \vect{AB}$ (red, here scaled by $\frac12$), and the two points used to build the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](#ex-g11-vect-twopoints) 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](#def-g11-vect-direction) $\vec u$ and $\vec v$ are parallel if and only if $\det(\vec u, \vec v) = 0$. In [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](#def-g11-vect-collinear) directions, which is [Theorem 15.5](#thm-g11-vect-collinearity); with $\vec u\,(-b, a)$ and $\vec v\,(-b', a')$, the [determinant](#def-g11-vect-det) is $(-b)a' - (-b')a = ab' - a'b$. If the lines are not parallel, solving the two [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) simultaneously (by substitution or combination) leads to a unique solution: the system behaves like a [nonzero-determinant](#def-g11-vect-det) $2 \times 2$ system. ∎

**Method 15.16 (Intersection of two lines).**

Solve the system of the two [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation): isolate one unknown in one [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) and substitute into the other, or combine the [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) to eliminate one unknown. Always check the resulting point in *both* [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation).

**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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $\vect{AB}$, $\vect{AC}$, $\vect{AB} + \vect{AC}$ and $2\vect{AB} -
3\vect{AC}$, and the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[BC]$.

**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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[BC]$ is $\left(\frac{5 + (-1)}{2}, \frac{3+1}{2}\right) = (2, 2)$.

**Exercise 15.2 ★.**

Are the [vectors](#def-g11-vect-vector) [collinear](#def-g11-vect-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 of Exercise 15.2.**

$\det = 4 \times 3 - (-2) \times (-6) = 12 - 12 = 0$: [collinear](#def-g11-vect-collinear) (indeed $\vec v = -\frac12 \vec u$).

$\det = 2 \times 7 - 3 \times 5 = -1 \neq 0$: not [collinear](#def-g11-vect-collinear).

$\det = 0 \times (-8) - 0 \times 3 = 0$: [collinear](#def-g11-vect-collinear) (both vertical).

**Exercise 15.3 ★.**

Find a cartesian [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the line through $A(2, 1)$ with [direction vector](#def-g11-vect-direction) $\vec u\,(3, -1)$, then of the line through $B(0, 4)$ and $C(2, 0)$.

**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](#def-g11-vect-direction), the [intersections](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) with the two coordinate axes, and decide whether the points $P(2, 6)$ and $Q(1, 4)$ belong to $d$.

**Solution of Exercise 15.4.**

A [direction vector](#def-g11-vect-direction) is $(-b, a) = (2, 3)$. [Intersections](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion): $x = 0$ 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 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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) 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 of Exercise 15.6.**

First pair: adding the two [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation), $(2x + y - 7) + (x - y + 1) =
3x - 6 = 0$, 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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion).

**Exercise 15.7 ★★.**

Give a parametric representation of the line $d: 2x - y + 3 = 0$, and a cartesian [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the line $\bigl\{(1 + 2t,\ -3 + t) : t \in \R\bigr\}$.

**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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation): $x - 2y - 7 = 0$ (check $t = 1$, point $(3, -2)$: $3 + 4 - 7 = 0$).

**Exercise 15.8 ★★.**

Find the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the *[median](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-median)* of the triangle $ABC$ issued from $C$ (the line joining $C$ to the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AB]$).

**Solution of Exercise 15.9.**

The [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AB]$ is $M\!\left(\frac{-1+3}{2}, \frac{0+2}{2}\right)
= (1, 1)$. The [median](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-median) is the line $(CM)$ with $C(1, -4)$: both points have [abscissa](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $1$, so the [median](https://one-course.com/books/math/2/en/chapter/8-descriptive-statistics#def-g10-stats-median) is the vertical line

$$
x - 1 = 0 .
$$

**Exercise 15.10 ★★.**

For which value(s) of the parameter $m$ are the [vectors](#def-g11-vect-vector) $\vec u\,(m, 2)$ and $\vec v\,(3, m - 1)$ [collinear](#def-g11-vect-collinear)?

**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](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) $m = \frac{1 \pm 5}{2}$, i.e. $m = 3$ or $m = -2$.

**Exercise 15.11 ★★★.**

Let $ABCD$ be a parallelogram, $I$ the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AB]$, and $J$ the point defined by $\vect{DJ} = \frac23 \vect{DI}$. Working in the [coordinate system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) where $A(0,0)$, $B(1,0)$, $D(0,1)$ (so $C(1,1)$):

1. Compute the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $I$ and $J$ .
2. 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 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](#def-g11-vect-parametric)), then turns to pure geometry with the other superpower of [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system): *choosing* them cleverly, as in [Exercise 15.11](#exo-g11-vect-11), to make classical theorems compute themselves.

**Part I — Lines, three ways.**

1. Give a parametric representation of the line through $A(1, 2)$ with [direction vector](#def-g11-vect-direction) $\vec u(3, -1)$ , then eliminate the parameter to obtain a Cartesian [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) .
2. For the line $2x - 5y + 3 = 0$ : read off a [direction vector](#def-g11-vect-direction) ( [Theorem 15.9](#thm-g11-vect-cartesian) ), find one point, and write a parametric representation.
3. Compute the [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) point of the two lines of questions 1 and 2 ( [Method 15.16](#met-g11-vect-intersection) ).
4. Confirm with the [determinant](#def-g11-vect-det) ( [Theorem 15.5](#thm-g11-vect-collinearity) ) that those two lines had to intersect; then show that $2x - 5y + 3 = 0$ and $-4x + 10y + 1 = 0$ are strictly parallel.
5. Are $A(2, 1)$ , $B(5, 3)$ , $C(11, 7)$ aligned ( [Method 15.7](#met-g11-vect-alignment) )?

**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).

6. Find the Cartesian [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the two *paths* , and compute where the paths cross.
7. 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.
8. 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](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#pb-g11-quad-1) vertex technology)?
9. 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.
10. Explain why, for *any* two ships on straight courses at constant speeds, $D^2(t)$ is always a [quadratic function](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-quadratic) of $t$ — so “closest point of approach” is always a vertex computation.
11. 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system).** Work in the frame of [Exercise 15.11](#exo-g11-vect-11): the parallelogram $ABCD$ becomes $A(0,0)$, $B(1,0)$, $C(1,1)$, $D(0,1)$; let $I$ be the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AB]$ and $K$ the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[CD]$.

12. Compute a Cartesian [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the line $(DI)$ .
13. Intersect $(DI)$ with the diagonal $(AC)$ : recover [Exercise 15.11](#exo-g11-vect-11) ’s fact — the cut lies exactly one third of the way up the diagonal.
14. Now the line $(BK)$ : find its [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) , intersect with $(AC)$ , and conclude the full classical theorem: *the cevians $(DI)$ and $(BK)$ trisect the diagonal $(AC)$* .
15. Why was it legitimate to prove the theorem in this one special frame? (What does the choice of [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) preserve, and what does the statement involve?) Answer in two or three sentences.
16. One more trisection for practice: let $E$ be the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[BC]$ . Where does the line $(AE)$ cut the *other* diagonal $(DB)$ ?

**Part IV — Position puzzles.**

17. Are the three lines $x + y = 4$ , $2x - y = 2$ , $x - 2y = -2$ concurrent? (Intersect two, test the third.)
18. 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](https://one-course.com/books/math/2/en/chapter/3-functions#pb-g10-functions-1) . (Group the terms in $m$ .)
19. In the family $(d_m)$ : which member is parallel to $y = 2x + 7$ ? Which passes through the origin?
20. Finale — the three costumes of a line: Cartesian (membership tests in one substitution), reduced ( [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) and [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) at sight), parametric (motion, timing, interception); the [determinant](#def-g11-vect-det) as parallelism oracle; and the well-chosen frame as a theorem factory. One sentence each.

**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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x + 3y - 7 = 0$.

**2.** For $ax + by + c = 0$, a [direction vector](#def-g11-vect-direction) 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](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#def-g11-quad-discriminant) $\frac{68 \pm \sqrt{24}}{20} = 3.4 \pm 0.245$: 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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) of $t$, so each difference of [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) is affine, and $D^2$ — a sum of two squared [affine functions](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) — 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint), alignment, and ratios along a given line (all expressible with [vectors](#def-g11-vect-vector) 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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion): $\frac x2 = 1 - x$: $x = \frac23$: the point $\left(\frac23, \frac13\right)$ — again a trisection point, now of the other diagonal. [Midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $m$.

**19.** Parallel to $y = 2x + 7$: [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $m = 2$, the line $2x - y - 4 = 0$. Through the origin: $2 - 3m = 0$: $m = \frac23$.

**20.** Cartesian: plug in a point, membership answered. Reduced: [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) and intercept visible, [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph) immediate. Parametric: the clock built in — collisions, closest approach, interception. The [determinant](#def-g11-vect-det): one multiplication cross, and parallelism is decided. The chosen frame: a parallelogram becomes a unit square, and a classical trisection theorem becomes three [intersections](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) of lines.