If the equation of the locus of a point equid | vedclass

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(A) Let the point be $(x, y)$. Since the point is equidistant from $(a_1, b_1)$ and $(a_2, b_2)$,we have: $(x - a_1)^2 + (y - b_1)^2 = (x - a_2)^2 + (

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$\frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)$

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(A) Let the point be $(x, y)$. Since the point is equidistant from $(a_1, b_1)$ and $(a_2, b_2)$,we have: $(x - a_1)^2 + (y - b_1)^2 = (x - a_2)^2 + (y - b_2)^2$ Expanding both sides: $x^2 - 2a_1x + a_1^2 + y^2 - 2b_1y + b_1^2 = x^2 - 2a_2x + a_2^2 + y^2 - 2b_2y + b_2^2$ Canceling $x^2$ and $y^2$ from both sides: $-2a_1x - 2b_1y + a_1^2 + b_1^2 = -2a_2x - 2b_2y + a_2^2 + b_2^2$ Rearranging the terms to the form $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$: $2(a_2 - a_1)x + 2(b_2 - b_1)y + (a_1^2 + b_1^2 - a_2^2 - b_2^2) = 0$ Multiplying by $-1$ to match the given form $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$: $2(a_1 - a_2)x + 2(b_1 - b_2)y + (a_2^2 + b_2^2 - a_1^2 - b_1^2) = 0$ Dividing by $2$: $(a_1 - a_2)x + (b_1 - b_2)y + \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) = 0$ Comparing this with the given equation,we get $c = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)$.

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

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