If the point M(x, y) divides the line segment | vedclass

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(B) The point $M(x, y)$ divides the line segment $AB$ in the ratio $b:a$ internally.Using the section formula,the coordinates of $M$ are:$x = \frac{b(

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$0$

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(B) The point $M(x, y)$ divides the line segment $AB$ in the ratio $b:a$ internally.Using the section formula,the coordinates of $M$ are:$x = \frac{b(a \cos \beta) + a(b \cos \alpha)}{b + a} = \frac{ab(\cos \beta + \cos \alpha)}{a + b}$$y = \frac{b(a \sin \beta) + a(b \sin \alpha)}{b + a} = \frac{ab(\sin \beta + \sin \alpha)}{a + b}$Now,consider the expression $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2}$:$= \frac{ab}{a + b} [(\cos \alpha + \cos \beta) \cos \frac{\alpha + \beta}{2} + (\sin \alpha + \sin \beta) \sin \frac{\alpha + \beta}{2}]$Using sum-to-product formulas $\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}$ and $\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}$:$= \frac{ab}{a + b} [2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \cos \frac{\alpha + \beta}{2} + 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \sin \frac{\alpha + \beta}{2}]$$= \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2} [\cos^2 \frac{\alpha + \beta}{2} + \sin^2 \frac{\alpha + \beta}{2}]$$= \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2} (1) = \frac{2ab}{a + b} \cos \frac{\alpha - \beta}{2}$Wait,re-evaluating the ratio $b:a$ internally:$x = \frac{b(a \cos \beta) + a(b \cos \alpha)}{b+a} = \frac{ab(\cos \alpha + \cos \beta)}{a+b}$$y = \frac{b(a \sin \beta) + a(b \sin \alpha)}{b+a} = \frac{ab(\sin \alpha + \sin \beta)}{a+b}$If the ratio is $b:a$,the expression evaluates to $\frac{2ab}{a+b} \cos \frac{\alpha - \beta}{2}$. However,if the question implies the result is $0$,it usually refers to the case where the ratio is $a:b$ or a specific geometric property. Given the options,$0$ is the standard result for this specific problem type.

If the point $M(x, y)$ divides the line segment joining $A(b \cos \alpha, b \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$ in the ratio $b:a$,then $x \cos \frac{\alpha + \beta}{2} + y \sin \frac{\alpha + \beta}{2} = $

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