A straight line through the origin bisects th | vedclass

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(A) Let the points be $A(a \cos \alpha, a \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.The midpoint $P$ of the line segment $AB$ is given by $P\l

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Perpendicular

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(A) Let the points be $A(a \cos \alpha, a \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.The midpoint $P$ of the line segment $AB$ is given by $P\left( \frac{a(\cos \alpha + \cos \beta)}{2}, \frac{a(\sin \alpha + \sin \beta)}{2} ight)$.The slope of the line $AB$ is $m_1 = \frac{a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha} = \frac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha} = \frac{2 \sin(\frac{\beta - \alpha}{2}) \cos(\frac{\beta + \alpha}{2})}{-2 \sin(\frac{\beta - \alpha}{2}) \sin(\frac{\beta + \alpha}{2})} = -\cot(\frac{\alpha + \beta}{2})$.The slope of the line $OP$ passing through the origin $(0,0)$ and $P$ is $m_2 = \frac{\frac{a(\sin \alpha + \sin \beta)}{2}}{\frac{a(\cos \alpha + \cos \beta)}{2}} = \frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{2 \sin(\frac{\alpha + \beta}{2}) \cos(\frac{\alpha - \beta}{2})}{2 \cos(\frac{\alpha + \beta}{2}) \cos(\frac{\alpha - \beta}{2})} = 	an(\frac{\alpha + \beta}{2})$.Now,$m_1 	imes m_2 = -\cot(\frac{\alpha + \beta}{2}) 	imes 	an(\frac{\alpha + \beta}{2}) = -1$.Since the product of the slopes is $-1$,the lines are perpendicular.

$A$ straight line through the origin bisects the line segment joining the points $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$. Then the lines are

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