If the coordinates of the points A, B, C, D a | vedclass

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(B) The midpoint $E$ of $AB$ is $\left( \frac{a + a'}{2}, \frac{b + b'}{2} \right)$.The midpoint $F$ of $CD$ is $\left( \frac{-a + a'}{2}, \frac{b - b

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$2ay - 2b'x = ab - a'b'$

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(B) The midpoint $E$ of $AB$ is $\left( \frac{a + a'}{2}, \frac{b + b'}{2} \right)$.The midpoint $F$ of $CD$ is $\left( \frac{-a + a'}{2}, \frac{b - b'}{2} \right)$.The slope $m$ of the line $EF$ is given by:$m = \frac{\frac{b - b'}{2} - \frac{b + b'}{2}}{\frac{a' - a}{2} - \frac{a' + a}{2}} = \frac{\frac{-2b'}{2}}{\frac{-2a}{2}} = \frac{b'}{a}$.The equation of the line passing through $E$ with slope $m$ is:$y - \frac{b + b'}{2} = \frac{b'}{a} \left( x - \frac{a + a'}{2} \right)$.Multiplying by $2a$:$2ay - a(b + b') = 2b'x - b'(a + a')$.$2ay - ab - ab' = 2b'x - ab' - a'b'$.$2ay - 2b'x = ab - a'b'$.

If the coordinates of the points $A, B, C, D$ are $(a, b), (a', b'), (-a, b)$ and $(a', -b')$ respectively,then the equation of the line bisecting the line segments $AB$ and $CD$ is

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