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If \(4 a^2+9 b^2-c^2+12 a b=0\), then the family of straight lines \(a x+b y+c=0\) is concurrent at
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The correct answer is:
\((2,3)\) or \((-2,-3)\)
Hint : \(\because 2 a+3 b-c=0\) or \(2 a+3 b+c=0\)
\(\begin{aligned}
& \Rightarrow c= \pm(2 a+3 b) \\
& \because \quad a x+b y+c=0 \\
& \Rightarrow \quad a x+b y \pm(2 a+3 b)=0 \\
& \Rightarrow \quad a(x \pm 2)+b(y \pm 3)=0 \text { (family of lines) } \\
& \Rightarrow \quad(-2,-3) \quad \text { or }(2,3)
\end{aligned}\)
\(\begin{aligned}
& \Rightarrow c= \pm(2 a+3 b) \\
& \because \quad a x+b y+c=0 \\
& \Rightarrow \quad a x+b y \pm(2 a+3 b)=0 \\
& \Rightarrow \quad a(x \pm 2)+b(y \pm 3)=0 \text { (family of lines) } \\
& \Rightarrow \quad(-2,-3) \quad \text { or }(2,3)
\end{aligned}\)
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