Fixed-base pencil and the parameter value

Text

Consider the pencil $(x-y+1)+k(2x+y-4)=0$.

• (a) Find the base point (support) of the pencil.
• (b) Find the value of $k$ for which the line of the pencil passes through $Q(3,0)$.

Solution

(a) The base point is common to all lines and is found by solving the system of the two generators: $\begin{cases}x-y+1=0\\ 2x+y-4=0\end{cases}$ Adding the equations: $3x-3=0\Rightarrow x=1$, hence $y=x+1=2$. The support is $S(1,2)$. (b) Impose passage through $Q(3,0)$: $(3-0+1)+k(6+0-4)=0\;\Rightarrow\; 4+2k=0\;\Rightarrow\; k=-2.$ $\boxed{\,S(1,2),\quad k=-2\,}$