The perpendiculars are drawn to lines $\mathrm{L}_1$ and $\mathrm{L}_2$ from the origin making an angle…

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
  1. $x^2-y^2+2 \sqrt{2} y+2=0$
  2. $x^2-y^2-2 \sqrt{2} y-2=0$
  3. $x^2-y^2+2 \sqrt{2} y-2=0$
  4. $x^2-y^2-2 \sqrt{2} y+2=0$

Solution

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 $\begin{aligned} & (x+y-\sqrt{2})(x-y+\sqrt{2})=0 \\ & \Rightarrow x^2-y^2+2 \sqrt{2} y-2=0 \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 1)

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