Which one of the following is the perpendicular form of the straight line \sqrt{3}x+2y=7?

  1. A. y=-\frac{\sqrt{3}}{2}x+\frac{7}{2}
  2. B. \frac{x}{(\frac{7}{\sqrt{3}})}+\frac{y}{(\frac{7}{2})}=1
  3. C. \frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=\sqrt{7}
  4. D. \frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=7

Correct Answer: C. \frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=\sqrt{7}

Explanation

To convert the equation Ax + By = C into the perpendicular (or normal) form x\cos\alpha + y\sin\alpha = p, divide the entire equation by \sqrt{A^2 + B^2}. Here, A=\sqrt{3} and B=2, so we divide by \sqrt{(\sqrt{3})^2 + 2^2} = \sqrt{3+4} = \sqrt{7}. This gives \frac{\sqrt{3}}{\sqrt{7}}x + \frac{2}{\sqrt{7}}y = \frac{7}{\sqrt{7}}. Simplifying the RHS yields \sqrt{7}.

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