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NDA 2 2025 Maths, question 58

Topic: Analytical geometry

Q58

The equation of the locus of a point equidistant from the points (a,b)(a,b) and (c,d)(c,d) is (a−c)x+(b−d)y+k=0(a-c)x+(b-d)y+k=0. What is the value of kk?

  1. (a) a2−c2+b2−d2a^{2}-c^{2}+b^{2}-d^{2}
  2. (b) c2+d2−a2−b2c^{2}+d^{2}-a^{2}-b^{2}
  3. (c) (a2−c2+b2−d2)/2(a^{2}-c^{2}+b^{2}-d^{2})/2
  4. (d) (c2+d2−a2−b2)/2(c^{2}+d^{2}-a^{2}-b^{2})/2Correct answer

Solution.

Answer: (d) (c2+d2−a2−b2)/2(c^{2}+d^{2}-a^{2}-b^{2})/2

  1. Equate squared distances: (x−a)²+(y−b)²=(x−c)²+(y−d)².
  2. Cancelling x²,y² and rearranging gives (a−c)x+(b−d)y+(c²+d²−a²−b²)/2=0.
  3. Compare constant terms.