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4 different pairs of lines are given in List I and the cosine of the angle between every pair of lines is given in List II.
Match the following:
The correct match is
$\text { A } \quad \text { B } \quad \text { C } \quad \text { D }$
Options:
Match the following:
The correct match is
$\text { A } \quad \text { B } \quad \text { C } \quad \text { D }$
Solution:
2299 Upvotes
Verified Answer
The correct answer is:
III I V IV
Take statement A with equation
$$
5 x^2+2 \sqrt{7} x y-y^2=0
$$
$$
\begin{aligned}
& \text { so, } \cos \theta=\frac{(\mathrm{a}+\mathrm{b})}{\sqrt{(\mathrm{a}-\mathrm{b})^2+4 \mathrm{~h}^2}}=\frac{(5-1)}{\sqrt{(5+1)^2+4 \times 7}} \\
& \cos \theta=\frac{4}{\sqrt{64}}=\frac{4}{8}=\frac{1}{2}
\end{aligned}
$$
Thus, this equation belongs to option (III)
Now, take $x^2+\sqrt{11 x y}+2 y^2=0$
so,
$$
\begin{aligned}
& \cos \theta=\frac{(\mathrm{a}+\mathrm{b})}{\sqrt{(\mathrm{a}-\mathrm{b})^2+4 \mathrm{~h}^2}}=\frac{(3+1)}{\sqrt{(3-1)^2+4 \times 2}}=\frac{4}{\sqrt{4+32}} \\
& \cos \theta=\frac{3}{\sqrt{42}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Thus, this equation belongs to option (I)
Take equation D: $3 x^2+4 \sqrt{2} x y+y^2=0$
now, $\cos \theta=\frac{(\mathrm{a}+\mathrm{b})}{\sqrt{(\mathrm{a}-\mathrm{b})^2+4 \mathrm{~h}^2}}=\frac{(3+1)}{\sqrt{(3-1)^2+4 \times 2}}=\frac{4}{\sqrt{4+32}}$
$$
\cos \theta=\frac{4}{\sqrt{36}}=\frac{4}{6}=\frac{2}{3}
$$
Thus, it belongs to option (IV)
Therefore option (c) is correct.
$$
5 x^2+2 \sqrt{7} x y-y^2=0
$$
$$
\begin{aligned}
& \text { so, } \cos \theta=\frac{(\mathrm{a}+\mathrm{b})}{\sqrt{(\mathrm{a}-\mathrm{b})^2+4 \mathrm{~h}^2}}=\frac{(5-1)}{\sqrt{(5+1)^2+4 \times 7}} \\
& \cos \theta=\frac{4}{\sqrt{64}}=\frac{4}{8}=\frac{1}{2}
\end{aligned}
$$
Thus, this equation belongs to option (III)
Now, take $x^2+\sqrt{11 x y}+2 y^2=0$
so,
$$
\begin{aligned}
& \cos \theta=\frac{(\mathrm{a}+\mathrm{b})}{\sqrt{(\mathrm{a}-\mathrm{b})^2+4 \mathrm{~h}^2}}=\frac{(3+1)}{\sqrt{(3-1)^2+4 \times 2}}=\frac{4}{\sqrt{4+32}} \\
& \cos \theta=\frac{3}{\sqrt{42}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$
Thus, this equation belongs to option (I)
Take equation D: $3 x^2+4 \sqrt{2} x y+y^2=0$
now, $\cos \theta=\frac{(\mathrm{a}+\mathrm{b})}{\sqrt{(\mathrm{a}-\mathrm{b})^2+4 \mathrm{~h}^2}}=\frac{(3+1)}{\sqrt{(3-1)^2+4 \times 2}}=\frac{4}{\sqrt{4+32}}$
$$
\cos \theta=\frac{4}{\sqrt{36}}=\frac{4}{6}=\frac{2}{3}
$$
Thus, it belongs to option (IV)
Therefore option (c) is correct.
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