IB Analysis and Approaches HL · Plane geometry: vectors and lines · Section 3 of 3

Equations of a line, intersections and distances

3 worked examples · By Álvaro López

Key idea

A line in the plane is described by a point and a direction, and from this idea come four ways of writing it: vector, parametric, symmetric and Cartesian. Knowing how to convert between them, deciding whether two lines meet, finding the distance from a point to a line and applying this to objects moving at constant velocity is typical material for Paper 1 and Paper 2.

Vector equation of a line

A line is fixed by a point that lies on it and by a direction. The point is given by its position vector $a$ (the vector from the origin $O$ to the point $A$ ), and the direction by a direction vector $b$ , which is any non-zero vector parallel to the line. To reach any point $R$ on the line, start at $A$ and move some multiple of the vector $b$ (forwards or backwards).

Line through A with direction vector b; an arbitrary point R has position vector r equal to a plus lambda b

In the plane it is written in components:

$(x y) = (a_{1} a_{2}) + λ (b_{1} b_{2})$

The vector equation is not unique: any point on the line can serve as $a$ , and any non-zero multiple of $b$ can serve as the direction vector. Two equations that look different can describe the same line.

If two points $A$ and $B$ of the line are known, a direction vector is $A B = b - a$ (vectors from the previous section: end minus start), and $A$ is taken as the point.

To check whether a point $P$ lies on the line, substitute its coordinates and check that the same $λ$ satisfies both components.

Parametric, symmetric and Cartesian equations

Separating the vector equation component by component gives the parametric equations:

${x y = a_{1} + λ b_{1} = a_{2} + λ b_{2}$

If $b_{1} \neq = 0$ and $b_{2} \neq = 0$ , solve for $λ$ in both and equate them: this is the symmetric equation,

$\frac{x - a _{1}}{b _{1}} = \frac{y - a _{2}}{b _{2}}$

Clearing denominators and rearranging gives the Cartesian equation (or general form) $a x + b y + d = 0$ . There is a useful relationship between its coefficients and those of the line.

The intuition: the points $(x, y)$ of the line satisfy $(x, y) \cdot (a, b)$ always equal to $- d$ , that is, their projection onto the direction of $(a, b)$ is constant. That is why the line is perpendicular to that vector.

Parallel and perpendicular lines

Two lines with direction vectors $b_{1}$ and $b_{2}$ and gradients $m_{1}$ and $m_{2}$ are related as follows.

The condition using the scalar (dot) product comes from the section on the scalar product: two non-zero vectors are perpendicular if and only if their scalar product is $0$ . For gradients it follows from this: with $b_{1} = (1, m_{1})$ and $b_{2} = (1, m_{2})$ , $b_{1} \cdot b_{2} = 1 + m_{1} m_{2} = 0$ .

A shortcut for writing related lines directly in Cartesian form: the line $a x + b y + d = 0$ is parallel to all lines of the form $a x + b y + d^{'} = 0$ (only the constant term changes), and it is perpendicular to those of the form $b x - a y + d^{'} = 0$ .

Intersection of two lines

In the plane, two lines can:

cross at a single point: direction vectors not proportional;
be distinct parallel lines: proportional direction vectors and no common point;
be coincident: proportional direction vectors and a common point.

Distance from a point to a line

The distance from a point $P$ to a line $ℓ$ is the smallest of the distances from $P$ to the points of $ℓ$ . It is reached at the foot $Q$ of the perpendicular to $ℓ$ drawn through $P$ : any other point of the line is further away, because $P Q$ is a leg of a right-angled triangle whose hypotenuse is the distance to that other point.

With the Cartesian equation there is a direct formula. To derive it, let $F$ be any point on the line $a x + b y + d = 0$ and $n = (a, b)$ its normal vector. The distance from $P$ is the length of the projection of $F P$ onto the direction of $n$ , that is, $\frac{∣ F P \cdot n ∣}{∣ n ∣}$ . Since $F$ satisfies $a x_{F} + b y_{F} = - d$ , the numerator is $∣ a x_{0} + b y_{0} + d ∣$ .

The distance between two parallel lines is found by taking any point on one of them and calculating its distance to the other. If the lines intersect, the distance between them is $0$ .

Application: motion at constant velocity

An object moving in a straight line at constant velocity traces out a line, and the vector equation captures this with time as the parameter:

Here the direction vector $v$ is not arbitrary: its modulus is the speed, and $t$ is no longer a free parameter but real time (in hours, seconds… depending on the units in the question). With speed $∣ v ∣$ in units of length per unit of time, the object covers $∣ v ∣ t$ units of length in a time $t$ .

Three typical questions:

Position at an instant: substitute $t$ .
Does it pass through a point? Solve $r_{0} + t v = point$ ; the $t$ must be the same in both components and also $t \geq 0$ .
Do two objects meet? They must be in the same place at the same time: $r_{A} (t) = r_{B} (t)$ , with the same $t$ for both. The paths (the lines) can cross even though the objects do not coincide, because they pass through the crossing point at different times.

To find the minimum distance between two objects, study the vector from one to the other,

$d (t) = r_{B} (t) - r_{A} (t) = (r_{0 B} - r_{0 A}) + t (v_{B} - v_{A})$

which is also a line (in time) with direction vector $v_{B} - v_{A}$ . There are two methods:

write $∣ d (t) ∣^{2}$ as a quadratic function of $t$ and find its minimum (the vertex of the parabola, or the derivative);
use the fact that the minimum occurs when $d (t)$ is perpendicular to $v_{B} - v_{A}$ , that is, $d (t) \cdot (v_{B} - v_{A}) = 0$ .

In both cases, if the minimum occurs at a time $t < 0$ , it is not valid: the smallest distance for $t \geq 0$ is the one at the initial instant.

Worked examples

Example 1 Basic

Line through two points

Consider the points $A (1, 3)$ and $B (5, 1)$ .

a) Find a direction vector of the line $A B$ whose components are coprime integers and whose first component is positive.

b) Find the Cartesian equation of the line $A B$ in the form $a x + b y + d = 0$ , with $a, b, d$ coprime integers and $a > 0$ .

c) The point $C (- 3, k)$ lies on the line $A B$ . Find $k$ .

Solution

a) $A B = B - A = (4 - 2)$ . Dividing by $2$ gives the direction vector $(2 - 1)$ . A vector equation is, for example,

$r = (13) + λ (2 - 1)$

b) A normal vector is $(1, 2)$ (swap the components of the direction vector and change the sign of one of them). The equation is $x + 2 y = d^{'}$ , and since $A$ is on the line, $d^{'} = 1 + 2 \cdot 3 = 7$ :

$x + 2 y - 7 = 0$

Check with $B$ : $5 + 2 \cdot 1 - 7 = 0$ .

c) Substitute $C$ : $- 3 + 2 k - 7 = 0$ , so $2 k = 10$ and $k = 5$ .

Example 2 Medium

A perpendicular and its point of intersection

The line $ℓ$ has vector equation $r = (- 1 4) + λ (3 - 2)$ . Consider the point $P (0, - 1)$ .

a) Find the Cartesian equation of $ℓ$ in the form $a x + b y + d = 0$ , with $a, b, d$ coprime integers and $a > 0$ .

b) Using the same convention, find the Cartesian equation of the line $m$ that passes through $P$ and is perpendicular to $ℓ$ .

c) Find the point $Q$ where $ℓ$ and $m$ intersect.

d) Calculate the distance from $P$ to $ℓ$ .

Solution

a) With $b = (3, - 2)$ a normal vector is $(2, 3)$ , so the equation is $2 x + 3 y = d^{'}$ . With the point $(- 1, 4)$ : $d^{'} = - 2 + 12 = 10$ .

$2 x + 3 y - 10 = 0$

b) The line $m$ has as direction vector a normal vector of $ℓ$ , $(2, 3)$ (check: $(3, - 2) \cdot (2, 3) = 6 - 6 = 0$ ). Its normal vector is therefore $(3, - 2)$ : $3 x - 2 y = d^{'}$ . With $P (0, - 1)$ : $d^{'} = 0 + 2 = 2$ .

$3 x - 2 y - 2 = 0$

c) Solve the system $2 x + 3 y = 10$ , $3 x - 2 y = 2$ . Multiplying the first by $2$ and the second by $3$ and adding: $4 x + 6 y + 9 x - 6 y = 20 + 6$ , that is, $13 x = 26$ and $x = 2$ . Then $3 y = 10 - 4 = 6$ , $y = 2$ . Hence $Q (2, 2)$ . Check in $m$ : $6 - 4 - 2 = 0$ .

d) As $Q$ is the foot of the perpendicular through $P$ , the distance is

$∣ P Q ∣ = ∣(2, 3)∣ = 13$

Using the formula: $\frac{∣ 2 \cdot 0 + 3 \cdot ( - 1 ) - 10 ∣}{2 ^{2} + 3 ^{2}} = \frac{13}{13} = 13$ . The two methods agree.

Example 3 Hard

Two boats: do they meet? How close do they get?

Two boats, $A$ and $B$ , sail in straight lines at constant velocity. Their positions, in kilometres relative to a lighthouse at the origin, $t$ hours after noon, are

$r_{A} = (12) + t (21) and r_{B} = (11 - 2) + t (- 2 2)$

a) Calculate the speed of $B$ , in km/h.

b) Find the point at which the paths of the two boats cross.

c) Find the time at which $A$ passes that point and the time at which $B$ passes it, and conclude whether the boats collide.

d) Find the time at which the distance between the boats is minimum and the exact value of that minimum distance.

Solution

a) The velocity of $B$ is $(- 2, 2)$ , so the speed is $(- 2)^{2} + 2^{2} = 8 = 22$ km/h.

b) For the paths, use different parameters: $s$ for $A$ and $u$ for $B$ .

${1 + 2 s 2 + s = 11 - 2 u = - 2 + 2 u$

From the second, $s = 2 u - 4$ . Substituting into the first: $1 + 4 u - 8 = 11 - 2 u$ , so $6 u = 18$ and $u = 3$ , $s = 2$ . The point is $r_{A} (2) = (5, 4)$ . Check with $B$ : $(11 - 6, - 2 + 6) = (5, 4)$ .

c) With the values above, $A$ passes through $(5, 4)$ at $t = 2$ and $B$ at $t = 3$ . The times are different, so the boats are not at that point at the same time. In fact they never meet: $r_{A} (t) = r_{B} (t)$ requires the same $t$ in both components, and the system $1 + 2 t = 11 - 2 t$ , $2 + t = - 2 + 2 t$ gives $t = \frac{5}{2}$ and $t = 4$ , which are incompatible. The paths cross, but the boats do not collide.

d) The vector from $A$ to $B$ is

$d (t) = r_{B} - r_{A} = (10 - 4 t t - 4)$

The minimum distance occurs when $d (t)$ is perpendicular to the relative velocity $v_{B} - v_{A} = (- 4, 1)$ :

$(10 - 4 t) (- 4) + (t - 4) \cdot 1 = 0 ⟹ 17 t = 44 ⟹ t = \frac{44}{17} h \approx 2.59 h$

To find the distance itself, $∣ d ∣^{2} = (10 - 4 t)^{2} + (t - 4)^{2} = 17 t^{2} - 88 t + 116$ . With $t = \frac{44}{17}$ :

$∣ d ∣^{2} = 116 - \frac{4 4 ^{2}}{17} = \frac{1972 - 1936}{17} = \frac{36}{17} ⟹ ∣ d ∣ = \frac{6}{17} = \frac{6 17}{17} km \approx 1.46 km$

(The parabola $17 t^{2} - 88 t + 116$ has its vertex at $t = \frac{88}{34} = \frac{44}{17}$ , which confirms the time.)

In the exam: what earns marks is using different parameters to cross the paths and then comparing the times: giving the point of intersection as a "meeting point" without that step is the common mistake. For the distance, set up $d (t)$ and write down the condition you impose (perpendicular to the relative velocity, or minimum of $∣ d ∣^{2}$ ).

Common mistakes

Summary

Line: $r = a + λ b$ (point $A$ with position vector $a$ and direction vector $b$ ). It is not unique: any point and any multiple of $b$ will do.
Parametric $x = a_{1} + λ b_{1}$ , $y = a_{2} + λ b_{2}$ ; symmetric $\frac{x - a _{1}}{b _{1}} = \frac{y - a _{2}}{b _{2}}$ ; Cartesian $a x + b y + d = 0$ , with normal $(a, b)$ and direction $(b, - a)$ , gradient $m = - \frac{a}{b}$ .
Parallel: proportional direction vectors (or $m_{1} = m_{2}$ ). Perpendicular: $b_{1} \cdot b_{2} = 0$ (or $m_{1} m_{2} = - 1$ ).
Intersection: equate with different parameters ( $λ$ , $μ$ ) and solve the system; proportional directions mean parallel or coincident lines.
Distance from $P (x_{0}, y_{0})$ to $a x + b y + d = 0$ : $\frac{∣ a x _{0} + b y _{0} + d ∣}{a ^{2} + b ^{2}}$ , or the length $P Q$ with $Q$ the foot of the perpendicular.
Motion: $r = r_{0} + t v$ ; speed $∣ v ∣$ . Two objects meet if they coincide at the same $t$ ; the minimum distance comes from $d (t) \cdot (v_{B} - v_{A}) = 0$ .

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