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

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(D) Let $(h, k)$ be a point on the locus. By the given condition,the distance from $(h, k)$ to $(a_1, b_1)$ is equal to the distance from $(h, k)$ to

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

Vedclass · Answer

(D) Let $(h, k)$ be a point on the locus. By the given condition,the distance from $(h, k)$ to $(a_1, b_1)$ is equal to the distance from $(h, k)$ to $(a_2, b_2)$.$(h - a_1)^2 + (k - b_1)^2 = (h - a_2)^2 + (k - b_2)^2$Expanding both sides:$h^2 - 2ha_1 + a_1^2 + k^2 - 2kb_1 + b_1^2 = h^2 - 2ha_2 + a_2^2 + k^2 - 2kb_2 + b_2^2$Simplifying the equation:$-2ha_1 + a_1^2 - 2kb_1 + b_1^2 = -2ha_2 + a_2^2 - 2kb_2 + b_2^2$$2h(a_2 - a_1) + 2k(b_2 - b_1) + a_1^2 + b_1^2 - a_2^2 - b_2^2 = 0$Multiplying by $-1$ to match the form $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$:$2h(a_1 - a_2) + 2k(b_1 - b_2) + a_2^2 + b_2^2 - a_1^2 - b_1^2 = 0$Dividing by $2$:$(a_1 - a_2)h + (b_1 - b_2)k + \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) = 0$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 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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