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

\frac{[(a_{2} + a_{1}) (a_{2} - a_{1}) + (b_{2} - b_{1}) (b_{2} + b_{1})]}{2}

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\frac{[(a_{2} + a_{1}) (a_{1} - a_{2}) + (b_{2} - b_{1}) (b_{1} + b_{2})]}{2}

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\frac{[(a_{2} + a_{1}) (a_{2} - a_{1}) + (b_{1} - b_{2}) (b_{2} + b_{1})]}{2}

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\frac{[(a_{2} + a_{1}) (a_{2} - a_{1}) + (b_{2} - b_{1}) (b_{2} + b_{1})]}{4}

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Solution

The correct option is A $\frac{[(a_{2} + a_{1}) (a_{2} - a_{1}) + (b_{2} - b_{1}) (b_{2} + b_{1})]}{2}$
From given, we have,

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

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

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

dividing the above equation by $2$ , we get,

$c = \frac{[a_{2}^{2} + b_{2}^{2} - a_{1}^{2} - b_{1}^{2}]}{2}$

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