$\quad$ If $4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0,$ the family of lines…
$\quad$ If $4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0,$ the family of lines $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is concurrent at one or the other of the two points-
$\left(-1,-\frac{1}{2}\right),(-2,-1)$
$(-1,-1),\left(-2,-\frac{1}{2}\right)$
$(-1,2),\left(\frac{1}{2},-1\right)$
$(1,2),\left(\frac{1}{2},-1\right)$
Solution
$
\text { } \begin{array}{l}
4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c \\
\quad \equiv(2 a+b)^{2}-3(2 a+b) c+2 c^{2}=0 \\
\quad \Rightarrow(2 a+b-2 c)(2 a+b-c)=0 \Rightarrow c=2 a+b \\
\quad \text { or } c=a+\frac{1}{2} b
\end{array}
$
The equation of the family of lines is
$
a(x+2)+b(y+1)=0 \text { or } a(x+1)+b\left(y+\frac{1}{2}\right)=0
$
giving the point of concurrence (-2,-1) or
$
\left(-1,-\frac{1}{2}\right)
$