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If the equation of the locus of a point equi-distant from the points $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + K = 0$ then the value of $K$ is

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$\sqrt{a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2}}$

$\frac{1}{2} (a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2})$

$a_{1}^{2} - a_{2}^{2} + b_{1}^{2} - b_{2}^{2}$

$\frac{1}{2} (a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2})$

answer is B.

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Detailed Solution

$A (a_{1}, b_{1}), B (a_{2}, b_{2}), P A = P B \Rightarrow P A^{2} = P B^{2}$

${(x - a_{1})}^{2} + {(y - b_{1})}^{2} = {(x - a_{2})}^{2} + {(y - b_{2})}^{2}$

$\Rightarrow 2 (a_{1} - a_{2}) x + 2 (b_{1} - b_{2}) y + a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2} = 0$

$(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + \frac{1}{2} (a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2}) = 0$

$k = \frac{1}{2} (a_{1}^{2} + b_{1}^{2} - a_{2}^{2} - b_{2}^{2})$

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