pencil of lines

Let

$A_{i}x+B_{i}y+C_i=0$(1)

be equations of some lines.  Use the short notations  $A_{i}x+B_{i}y+C_i:=L_i$.

If the lines  $L_1=0$  and  $L_2=0$  have an intersection point $P$, then, by the parent entry (http://planetmath.org/LineThroughAnIntersectionPoint), the equation

$k_1{L}_1+k_2{L}_2=0$(2)

with various real values of $k_1$ and $k_2$ can any line passing through the point $P$; this set of lines is called a pencil of lines.

Theorem.  A necessary and sufficient condition in to three lines

$$L_1=0,L_2=0,L_3=0$$

pass through a same point, is that the determinant formed by the coefficients of their equations (1) vanishes:

Proof.  If the line  $L_3=0$  belongs to the fan of lines determined by the lines  $L_1=0$  and  $L_2=0$,  i.e. all the three lines have a common point, there must be the identity

$$L_3\equiv L_1+L_2,$$

i.e. there exist three real numbers $k_1$, $k_2$, $k_3$, which are not all zeroes, such that the equation

$k_1{L}_1+k_2{L}_2+k_3{L}_3\equiv 0$(3)

is satisfied identically by all real values of $x$ and $y$.This means that the group of homogeneous linear equations

$\{\begin{array}{cc}{k}_1{A}_1+k_2{A}_2+k_3{A}_3=0,\hfill & \\ k_1{B}_1+k_2{B}_2+k_3{B}_3=0,\hfill & \\ k_1{C}_1+k_2{C}_2+k_3{C}_3=0\hfill & \end{array}$

has nontrivial solutions $k_1,k_2,k_3$.By linear algebra, it follows that the determinant of this group of equations has to vanish.

Suppose conversely that the determinant vanishes.  This implies that the above group of equations has a nontrivial solution $k_1,k_2,k_3$.  Thus we can write the identic equation (3).  Let e.g.  $k_1\ne 0$.  Solving (3) for $L_1$ yields

$$L_1\equiv -\frac{k_2{L}_2+k_3{L}_3}{k_1},$$

which shows that the line  $L_1=0$  belongs to the fan determined by the lines  $L_2=0$  and  $L_3=0$; so the lines pass through a common point.