Q127 / 225 · JEE Main 2003 · Mathematics

If the equation of the locus of a point equidistant from the point $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right)$ is $\left(a_1-b_2\right) x+\left(a_1-b_2\right) y+c=0$, then the value of 'c' is

  • A
    $\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2-\mathrm{a}_2^2-\mathrm{b}_2^2}$
  • Correct answer
    B
    $\frac{1}{2} a_2^2+b_2^2-a_1^2-b_1^2$
  • C
    $\mathrm{a}_1^2-\mathrm{a}_2^2+\mathrm{b}_1^2-\mathrm{b}_2^2$
  • D
    $\frac{1}{2}\left(a_1^2+a_2^2+b_1^2+b_2^2\right)$

Solution

$\left(h-a_1\right)^2+\left(k-b_1\right)^2=\left(h-a_2\right)^2+\left(k-b_2\right)^2$ $\left(a_1-a_2\right) x+\left(b_1-b_2\right) y+\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)=0$ $C=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)$

Hence, the correct answer is option: 'B'.

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