Fixed-base pencil and the parameter value

Text

Given the pencil $(2x+y-5)+k(x-y-1)=0$, find the base point of the pencil and the pencil line through $Q(4,0)$.

Solution

The base point is the intersection of the two generators: from $2x+y=5$ and $x-y=1$, adding $3x=6\Rightarrow x=2,\ y=1$, i.e. $S(2,1)$. Impose passage through $Q(4,0)$: $(2\cdot4+0-5)+k(4-0-1)=0\Rightarrow 3+3k=0\Rightarrow k=-1.$ Substituting: $(2x+y-5)-(x-y-1)=x+2y-4=0$. $\boxed{\,S(2,1),\quad k=-1,\quad x+2y-4=0\,}$