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

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Question

(a) Let the point be $(h,\,k)$
${(h - {a_1})^2} + {(k - {b_1})^2} = {(h - {a_2})^2} + {(k - {b_2})^2}$
Replace $(h,\,k)$ by $(x,\,y)$, we get
$({a_

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) Let the point be $(h,\,k)$
${(h - {a_1})^2} + {(k - {b_1})^2} = {(h - {a_2})^2} + {(k - {b_2})^2}$
Replace $(h,\,k)$ by $(x,\,y)$, we get
$({a_1} - {a_2})x + ({b_1} - {b_2})y + \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2) = 0$
$c = \frac{1}{2}(a_2^2 + b_2^2 - a_1^2 - b_1^2)$.

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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