Pencils of lines: proper and improper

End of September 2026: in the technical track and at the science high school, the chapter on the line in the Cartesian plane — gradient (slope), parallelism, perpendicularity, point-to-line distance — closes in mid-to-late October with the pencils of lines (families of lines described by a parameter), the last topic before the test. The test comes between the end of October and mid-November: from here you have four to seven weeks. If the rest of the chapter on the line is already solid, pencils are the quickest topic to catch up on in the whole chapter, because it is always the same idea applied to different data.

Here you will find what distinguishes a proper pencil from an improper one, how to find the centre of a pencil, the method for finding the line of a pencil that satisfies an extra condition, the typical mistakes and three exercises worked in full.

Improper pencil: lines that are all parallel

The simplest case is the improper pencil: the set of all lines that have the same gradient $m$ , so all parallel to each other. It is written by fixing $m$ and letting the constant term vary:

y = m x + k

As the parameter $k \in R$ varies you obtain all and only the lines parallel to $y = m x$ , each shifted vertically by a different amount. Two lines of this pencil never meet — it is the very definition of parallel lines — and that is why it is called “improper”: the pencil has no real centre, in the sense that their point of intersection is an improper point (at infinity), a concept that as a first approximation you can simply read as “does not exist”.

Proper pencil: lines through the same point

The proper pencil is the set of all lines passing through the same point $C (x_{0}, y_{0})$ , called the centre of the pencil. The most direct way to write it is the point-slope form with $m$ as the parameter:

y - y_{0} = m (x - x_{0})

As $m \in R$ varies you obtain all the lines through $C$ except one: the vertical line $x = x_{0}$ , which has no gradient and so never appears in this form, for any value of $m$ . It is the only line of the proper pencil that has to be added separately, when needed, by writing it explicitly as $x = x_{0}$ .

The pencil as a combination of two generating lines

There is a second way to obtain a proper pencil, useful when the centre is not given directly but is defined as the intersection of two lines. Given two generating lines

r : a x + b y + c = 0 r^{'} : a^{'} x + b^{'} y + c^{'} = 0

not parallel to each other, the set

r + k \cdot r^{'} = 0, k \in R

is the proper pencil with centre at the point of intersection of $r$ and $r^{'}$ — even if that point has never been calculated explicitly. The reason is simple: if a point $P$ satisfies both $r = 0$ and $r^{'} = 0$ , it also satisfies $r + k \cdot r^{'} = 0$ for any $k$ , because it is $0 + k \cdot 0$ . The centre of the pencil is therefore found by putting the two generating lines into a system, exactly as you find the intersection of any two lines.

Here the exclusion is different from that of the point-slope form: in $r + k \cdot r^{'} = 0$ the line that is not obtained for any finite value of $k$ is the generating line $r^{'}$ itself (you would “reach” it only for $k \to \infty$ ). It is not necessarily a vertical line: it depends on how $r$ and $r^{'}$ are written. Two different exclusions, two different reasons: it is one of the points on which the two forms of the proper pencil get confused.

Finding the line of the pencil that you need

The test rarely asks only “write the pencil”: it asks for the line of the pencil with an extra property. The method is always the same: write the pencil with a parameter (usually $k$ ), impose the required condition, solve for the parameter, substitute.

It passes through a point $P$ : substitute the coordinates of $P$ for $x$ and $y$ in the equation of the pencil and solve for $k$ .
It is parallel to a given line $s$ : require the gradient of the line of the pencil to equal that of $s$ .
It is perpendicular to a given line $s$ : require the product of the two gradients to be $- 1$ (a note: in test practice this condition comes up less often than the parallelism one, but the procedure is identical).
It has a given distance from a point: use the point-to-line distance formula, requiring it to equal the requested number; it is the longest case, and generally appears only in the more complete tests.

The mistakes that cost marks

1. Forgetting the excluded line. If the pencil is written as $y - y_{0} = m (x - x_{0})$ and the exercise asks for “all the lines of the pencil with a certain property”, a complete answer also checks whether $x = x_{0}$ satisfies that property, because that line does not appear among the values of $m$ .

2. Confusing proper and improper pencils when the generating lines are parallel. If you choose two generating lines that are parallel to each other, the “pencil” $r + k \cdot r^{'} = 0$ has no real centre: it is in fact an improper pencil (all the lines it produces are parallel to the generating lines), even though the notation looks like that of a proper pencil. Before applying the method, always check that the two generating lines are not parallel — that is, that their gradients are different, or equivalently that the system between the two has exactly one solution.

3. Getting the sign of $k$ wrong. It is the most frequent mistake of all: solving the equation in the parameter with an inverted sign leads to a line that does not pass through the required point or does not have the required gradient. After finding $k$ , it is always worth a quick check: substitute the point or verify the gradient on the line you found.

4. Losing the centre in the step to $k \to \infty$ . The centre of the pencil satisfies both $r = 0$ and $r^{'} = 0$ at the same time, so it satisfies every line of the pencil, including the one you have just found by imposing a condition. If after calculating the line you want you check that the centre does not lie on it, you have made a calculation error somewhere: it is the same consistency check used for tangents to the parabola.

Three worked examples

1. Finding the centre and the line of the pencil through a point

Given the pencil generated by $r : x + y - 3 = 0$ and $r^{'} : x - y - 1 = 0$ , find the centre and the line of the pencil passing through $P (0, 5)$ .

The centre is found by putting the two generating lines into a system:

{x + y - 3 = 0 x - y - 1 = 0

Adding the two equations eliminates $y$ : $2 x - 4 = 0 \Rightarrow x = 2$ , and substituting $y = 1$ . The centre is $C (2, 1)$ .

The pencil is written $(x + y - 3) + k (x - y - 1) = 0$ . I impose passing through $P (0, 5)$ , substituting its coordinates:

(0 + 5 - 3) + k (0 - 5 - 1) = 0 \Rightarrow 2 - 6 k = 0 \Rightarrow k = \frac{1}{3}

I substitute into the pencil and multiply by 3 to remove the fraction:

(x + y - 3) + \frac{1}{3} (x - y - 1) = 0 ⟹ 3 (x + y - 3) + (x - y - 1) = 0 ⟹ 4 x + 2 y - 10 = 0

Simplifying by 2: $2 x + y - 5 = 0$ , that is $y = - 2 x + 5$ .

Consistency check: the centre $C (2, 1)$ must lie on this line. $2 \cdot 2 + 1 - 5 = 0$ . It works. And $P (0, 5)$ : $0 + 5 - 5 = 0$ . That works too.

2. Line of the pencil parallel to a given line

With the same pencil as in the previous exercise — generating lines $r : x + y - 3 = 0$ and $r^{'} : x - y - 1 = 0$ , centre $C (2, 1)$ — find the line of the pencil parallel to $s : 3 x - y + 2 = 0$ (gradient $3$ ).

I write the pencil in general form, collecting the terms in $x$ and $y$ :

(x + y - 3) + k (x - y - 1) = 0 ⟹ (1 + k) x + (1 - k) y - 3 - k = 0

The gradient of this line (for $1 - k \neq = 0$ ) is

m = - \frac{1 + k}{1 - k}

I impose $m = 3$ :

- \frac{1 + k}{1 - k} = 3 ⟹ - (1 + k) = 3 (1 - k) ⟹ - 1 - k = 3 - 3 k ⟹ 2 k = 4 ⟹ k = 2

I substitute $k = 2$ into the general equation:

(1 + 2) x + (1 - 2) y - 3 - 2 = 0 ⟹ 3 x - y - 5 = 0 ⟹ y = 3 x - 5

Checks: the gradient is $3$ , as required, and the line does not coincide with $s$ (which has a different constant term, $+ 2$ against $- 5$ : they are distinct parallel lines, not the same line). The centre $C (2, 1)$ : $3 \cdot 2 - 1 - 5 = 0$ . It works.

3. Improper pencil through a point

Find the line parallel to $y = 2 x - 7$ that passes through $A (3, 2)$ .

The lines parallel to $y = 2 x$ form the improper pencil $y = 2 x + k$ . I impose passing through $A (3, 2)$ :

2 = 2 \cdot 3 + k ⟹ 2 = 6 + k ⟹ k = - 4

The line sought is $y = 2 x - 4$ . Check: $2 \cdot 3 - 4 = 2$ , the point lies on the line, and the gradient is $2$ like that of the given line. Before starting, though, always check that the point does not already lie on the given line: if it does, the “parallel through $A$ ” is the line itself, and the question probably has a typo or a trap.

From here onwards

The procedure you have just used — pencil, parameter, condition, substitution — is the same one you will meet again in Year 3 with pencils of parabolas, where in place of two generating lines there are two parabolas (or a parabola and a line) and the parameter also serves to tell when the pencil degenerates from a parabola into a line. I wrote about it in the parabola in Year 3: tangents, pencils and problems with a parameter, if you want to see where it leads.

If the rest of the chapter on the line — gradient, point-to-line distance, line through two points — is not yet solid, start from there: I wrote it up in full in factorisation and the line: the first Year 2 test. And if in class the line is intertwined with linear systems (to find the intersection of two given lines, not generic ones as in the pencil), the dedicated guide is linear systems and the line in Year 2.

To train on closed-answer questions on the whole chapter — gradient, parallel and perpendicular lines, point-to-line distance and pencils — there is the mock test on the line in the Cartesian plane: 15 questions in 50 minutes, with a breakdown of weak points by topic. The correct answer to every question is free and without registration; the guided step-by-step solution unlocks with the first lesson.

If after the test pencils remain the weak point, or if it is the whole chapter on the line that is not working for you, the free introductory call is for exactly that: you tell me what you have done in class and I tell you honestly where to concentrate the first lessons before the test. The maths lessons are online and one-to-one: we start with a free 30-minute introductory call.

Pencils of lines: proper and improper, worked examples