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

  1. A. x\sin \theta-y\cos \theta-1=0
  2. B. x\sin \theta-y\cos \theta+1=0 ✓
  3. C. x\sin \theta-y\cos \theta=0
  4. D. x\cos \theta-y\sin \theta+1=0

Correct Answer: B. x\sin \theta-y\cos \theta+1=0

Explanation

The slope of the given line x\cos \theta+y\sin \theta=9 is -\frac{\cos \theta}{\sin \theta} = -\cot \theta. The slope of the line perpendicular to it is \tan \theta = \frac{\sin \theta}{\cos \theta}. The equation of the required line is y - \cos \theta = \frac{\sin \theta}{\cos \theta}(x - (-\sin \theta)). Multiplying by \cos \theta: y\cos \theta - \cos^2 \theta = x\sin \theta + \sin^2 \theta. Rearranging terms, x\sin \theta - y\cos \theta + (\sin^2 \theta + \cos^2 \theta) = 0, which simplifies to x\sin \theta - y\cos \theta + 1 = 0.

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