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The value of $\lambda$ such that $\lambda x^2-10 x y+12 y^2+5 x-16 y-3=0$ represents a pair of straight lines, is
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Verified Answer
The correct answer is:
$2$
Given pair of lines is
$$
\lambda x^2-10 x y+12 y^2+5 x-16 y-3=0
$$
On comparing with
$$
a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
$$
we get,
$$
a=\lambda, h=-5, b=12, g=\frac{5}{2}, f=-8, c=-3
$$
Condition for represents a pair of lines is
$$
\begin{aligned}
& a b c+2 f g h-a f^2-b g^2-c h^2=0 \\
& \lambda \times 12 \times(-3)+2(-8)\left(\frac{5}{2}\right)(-5)-\lambda(-8)^2 \\
& -12\left(\frac{5}{2}\right)^2+3(-5)^2=0 \\
& \Rightarrow-36 \lambda+200-64 \lambda-75+75=0 \\
& \Rightarrow \quad 100 \lambda=200 \Rightarrow \lambda=2 \\
&
\end{aligned}
$$
$$
\lambda x^2-10 x y+12 y^2+5 x-16 y-3=0
$$
On comparing with
$$
a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
$$
we get,
$$
a=\lambda, h=-5, b=12, g=\frac{5}{2}, f=-8, c=-3
$$
Condition for represents a pair of lines is
$$
\begin{aligned}
& a b c+2 f g h-a f^2-b g^2-c h^2=0 \\
& \lambda \times 12 \times(-3)+2(-8)\left(\frac{5}{2}\right)(-5)-\lambda(-8)^2 \\
& -12\left(\frac{5}{2}\right)^2+3(-5)^2=0 \\
& \Rightarrow-36 \lambda+200-64 \lambda-75+75=0 \\
& \Rightarrow \quad 100 \lambda=200 \Rightarrow \lambda=2 \\
&
\end{aligned}
$$
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