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

  1. A. a^{2}-c^{2}+b^{2}-d^{2}
  2. B. c^{2}+d^{2}-a^{2}-b^{2}
  3. C. (a^{2}-c^{2}+b^{2}-d^{2})/2
  4. D. (c^{2}+d^{2}-a^{2}-b^{2})/2 ✓

Correct Answer: D. (c^{2}+d^{2}-a^{2}-b^{2})/2

Explanation

Let the point be (x, y). Equating the squared distances: (x-a)^2 + (y-b)^2 = (x-c)^2 + (y-d)^2. Expanding and cancelling x^2 and y^2 gives -2ax + a^2 - 2by + b^2 = -2cx + c^2 - 2dy + d^2. Rearranging into the required form gives 2(a-c)x + 2(b-d)y + (c^2+d^2-a^2-b^2) = 0. Dividing by 2 yields k = \frac{c^2+d^2-a^2-b^2}{2}.

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