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If the equation of the locus of a point equidistant from the point $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right)$ is $\left(a_1-b_2\right) x+\left(a_1-b_2\right) y+c=0$, then the value of 'c' is
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The correct answer is:
$\frac{1}{2} a_2^2+b_2^2-a_1^2-b_1^2$
$\frac{1}{2} a_2^2+b_2^2-a_1^2-b_1^2$
$\left(h-a_1\right)^2+\left(k-b_1\right)^2=\left(h-a_2\right)^2+\left(k-b_2\right)^2$
$\left(a_1-a_2\right) x+\left(b_1-b_2\right) y+\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)=0$
$C=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)$
$\left(a_1-a_2\right) x+\left(b_1-b_2\right) y+\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)=0$
$C=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)$
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