The lines x α + y α = p₁ and x β + y β = p₂ will be perpendicular if

Correct Answer is : ∣α−β∣=π2|α - β| = π/2∣α−β∣=2π

Correct Answer is : ∣α−β∣=π2|α - β| = π/2∣α−β∣=2π

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UPSC NDA Math Model Paper 2

Question : 115 of 120

Marks: +1, -0

The lines

x \cos \alpha + y \sin \alpha = p_{1}

x \cos \beta + y \sin \beta = p_{2}

will be perpendicular if

Solution:

x \cos \alpha + y \sin \alpha = p_{1}

\Rightarrow y \sin \alpha = -x \cos \alpha + p_{1}

\Rightarrow y = -x \frac{\cos \alpha}{\sin \alpha} + \frac{p_{1}}{\sin \alpha}

and

x \cos \beta + y \sin \beta = p_{2}

\Rightarrow y \sin \beta = -\cos \beta + p_{2}

\Rightarrow y = -x \frac{\cos \beta}{\sin \beta} + \frac{p_{2}}{\sin \beta}

These lines are perpendicular so, product of slopes

= -1

\left(-\frac{\cos \alpha}{\sin \alpha}\right) \times \left(-\frac{\cos \beta}{\sin \beta}\right) = -1

\cos \alpha \cos \beta + \sin \alpha \sin \beta = 0

\cos (\alpha - \beta) = 0

\cos (\alpha - \beta) = \cos \left( \frac{\pi}{2} \right)

\therefore |\alpha - \beta| = \frac{\pi}{2}

Hence, option (2) is correct.

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