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NDA 2 2025 Maths, question 61

Topic: Analytical geometry

Q61

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

  1. (a) y=−32x+72y=-\frac{\sqrt{3}}{2}x+\frac{7}{2}
  2. (b) x(73)+y(72)=1\frac{x}{\left( \frac{7}{\sqrt{3}} \right)}+\frac{y}{\left( \frac{7}{2} \right)}=1
  3. (c) 37x+27y=7\frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=\sqrt{7}Correct answer
  4. (d) 37x+27y=7\frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=7

Solution.

Answer: (c) 37x+27y=7\frac{\sqrt{3}}{\sqrt{7}}x+\frac{2}{\sqrt{7}}y=\sqrt{7}

  1. Normal or perpendicular form has a unit normal.
  2. Divide √3x+2y=7 by √((√3)²+2²)=√7.
  3. This gives (√3/√7)x+(2/√7)y=√7, and the squared normal coefficients add to 1.