The single-parameter notation

Normalising $\alpha=1$ in the two-generator form gives the single-parameter notation used in practice: $r_1 + k\,r_2 = 0,$ where $k$ is the parameter of the pencil. One must remember, however, that this normalisation excludes one of the two generators: precisely, letting $k\to\infty$ one formally “obtains” $r_2=0$, but no finite value of $k$ gives $r_2$.

Warning

When solving problems with $r_1 + k\,r_2 = 0$, one must always check separately whether the generator $r_2$ is itself a solution of the problem: it is never reached by a finite value of $k$.

Remark — How to recognise the generators of a pencil written as $r_1+k\,r_2=0$

If a pencil is given in the form $F(x,y) + k\,G(x,y) = 0$, the two generators are simply $F=0$ and $G=0$: one just collects everything that does not depend on $k$ into $F$, and everything that multiplies $k$ into $G$. If then $F$ and $G$ are incident, the pencil is proper; if they are parallel, it is improper.

Example — Finding the support of a proper pencil

Given the pencil $f_k:\ x - ky + k + 2 = 0$, determine the support (if the pencil is proper).

Collecting the parameter: $k\,(-y + 1) + (x + 2) = 0$.

The two generators are $r_1: x+2 = 0$ (that is $x=-2$) and $r_2: -y+1=0$ (that is $y=1$). They are two incident lines (one vertical and one horizontal), so the pencil is proper with support at their point of intersection $P(-2;1)$.

In the following simulation the proper pencil is generated by $r_1:\ x - y - 1 = 0$ and $r_2:\ x + y - 3 = 0$: as the parameter $k$ varies, the line $r_1 + k\,r_2 = 0$ rotates about the support $P(2;1)$.

Drag the slider $k$: every line of the pencil (in red) passes through the support $P(2;1)$, but no finite value of $k$ returns the generator $r_2$.

Links

Topics: Pencils of lines
Concepts: Pencil of lines · Proper pencil · Generators · Parameter of the pencil · Support of the pencil
Functions: Line
Skills: Analytic geometry · Solving systems