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$\quad$ 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-
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Verified Answer
The correct answer is:
$\left(-1,-\frac{1}{2}\right),(-2,-1)$
$$
\text { } \begin{array}{l}
4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c \\
\quad \equiv(2 a+b)^{2}-3(2 a+b) c+2 c^{2}=0 \\
\quad \Rightarrow(2 a+b-2 c)(2 a+b-c)=0 \Rightarrow c=2 a+b \\
\quad \text { or } c=a+\frac{1}{2} b
\end{array}
$$
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
$$
\left(-1,-\frac{1}{2}\right)
$$
\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)
$$
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