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NDA 1 2024 Maths, question 59

Topic: Analytical geometry

Q59

What is the equation to the straight line passing through the point (−sin⁡θ,cos⁡θ)(-\sin \theta,\cos \theta) and perpendicular to the line xcos⁡θ+ysin⁡θ=9x\cos \theta+y\sin \theta=9 ?

  1. (a) xsin⁡θ−ycos⁡θ−1=0x\sin \theta-y\cos \theta-1=0
  2. (b) xsin⁡θ−ycos⁡θ+1=0x\sin \theta-y\cos \theta+1=0Correct answer
  3. (c) xsin⁡θ−ycos⁡θ=0x\sin \theta-y\cos \theta=0
  4. (d) xcos⁡θ−ysin⁡θ+1=0x\cos \theta-y\sin \theta+1=0

Solution.

Answer: (b) xsin⁡θ−ycos⁡θ+1=0x\sin \theta-y\cos \theta+1=0

  1. A perpendicular direction gives normal (sin θ, −cos θ).
  2. Substituting the required point (−sin θ, cos θ) fixes the constant as 1, because sin²θ + cos²θ = 1.
  3. Hence the line is x sin θ − y cos θ + 1 = 0.