MHT CET · Maths · Pair of Lines

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

A \(x^2-y^2+2 \sqrt{2} y+2=0\)
B \(x^2-y^2-2 \sqrt{2} y-2=0\)
C \(x^2-y^2+2 \sqrt{2} y-2=0\)
D \(x^2-y^2-2 \sqrt{2} y+2=0\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(x^2-y^2+2 \sqrt{2} y-2=0\)

Step-by-step Solution

Detailed explanation

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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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