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The perpendiculars are drawn to lines $\mathrm{L}_1$ and $\mathrm{L}_2$ from the origin making an angle $\frac{\pi}{4}$ and $\frac{3 \pi}{4}$ respectively with positive direction of $\mathrm{X}$-axis. If both the lines are at unit distance from the origin, then their joint equation is
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Verified Answer
The correct answer is:
$x^2-y^2+2 \sqrt{2} y-2=0$
Equation of line $\mathrm{L}_1$ is
$$
\begin{aligned}
& x \cos \frac{\pi}{4}+y \sin \frac{\pi}{4}=1 \\
& \Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=1 \\
& \Rightarrow x+y-\sqrt{2}=0
\end{aligned}
$$
Equation of line $L_2$ is
$$
\begin{aligned}
& x \cos \frac{3 \pi}{4}+y \sin \frac{3 \pi}{4}=1 \\
& \Rightarrow \frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=1 \\
& \Rightarrow x-y+\sqrt{2}=0
\end{aligned}
$$
$\therefore \quad$ The joint equation of the lines is
$$
\begin{aligned}
& (x+y-\sqrt{2})(x-y+\sqrt{2})=0 \\
& \Rightarrow x^2-y^2+2 \sqrt{2} y-2=0
\end{aligned}
$$
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