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

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

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

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

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

answer is A.

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Detailed Solution

$A (a_{1}, b_{1}), B (a_{2}, b_{2}), P (x, y) PA = PB$

$Let (x, y) be equidistant from (a_{1}, b_{1}) and (a_{2}, b_{2}) Then using distance formula {(a_{1} - x)}^{2} + {(b_{1} - y)}^{2} = {(a_{2} - x)}^{2} + {(b_{2} - y)}^{2} solving above, we will get 2 (a_{1} - a_{2}) x + 2 (b_{1} - b_{2}) y + [a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2}] Divide whole equation by 2 a n d on comparing with (a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0 \Rightarrow c = \frac{a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2}}{2}$