Co-ordinates of the foot of the perpendicular drawn from \( \Large \left(0,\ 0\right) \) to the line joining \( \Large \left(a \cos \alpha ,\ a \sin \alpha\right)\ and\ \left(a \cos \beta ,\ a \sin \beta \right) \) are:


A) \( \Large \left(\frac{a}{2},\ \frac{b}{2}\right) \)

B) \( \Large \left(\frac{a}{2} \left(\cos \alpha + \cos \beta \right), \frac{a}{2} \left(\sin \alpha + \sin \beta \right) \right) \)

C) \( \Large \left(\cos\frac{ \alpha + \beta }{2},\ \sin\frac{ \alpha + \beta }{2}\right) \)

D) \( \Large \left(0,\ \frac{b}{2}\right) \)

Correct Answer:
B) \( \Large \left(\frac{a}{2} \left(\cos \alpha + \cos \beta \right), \frac{a}{2} \left(\sin \alpha + \sin \beta \right) \right) \)

Description for Correct answer:

Slope of perpendicular to the line joining the points \( \Large \left(a,\ \cos \alpha ,\ a\ \sin \alpha \right)\ and\ \left(a\ \cos \beta ,\ a\ \sin \beta \right) \)

= \( \Large \frac{-\cos \alpha -\cos \beta }{\sin \alpha -\ \sin \beta } = \tan \left(\frac{ \alpha + \beta }{2} \right) \)

Hence, equation of perpendicular is

\( \Large y\ =\ \tan \left(\frac{ \alpha + \beta }{2}\right)x \) ...(i)

Now, on solving the equation of line with equation (i) we get

\( \Large \left[ \frac{a}{2} \left(cos \alpha +\ \cos \beta \right),\ \frac{a}{2} \left(\sin \alpha + \sin \beta \right) \right] \)


Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products








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