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

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Question

(A) Let $(x, y)$ be any point on the locus.By the condition of equidistance,the distance from $(x, y)$ to $(a_1, b_1)$ equals the distance to $(a_2, b

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $(x, y)$ be any point on the locus.By the condition of equidistance,the distance from $(x, y)$ to $(a_1, b_1)$ equals the distance to $(a_2, b_2)$.$(x - a_1)^2 + (y - b_1)^2 = (x - a_2)^2 + (y - b_2)^2$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 match the 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 the entire equation 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 $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,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 $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,find the value of $c$.

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