Q. Coordinate of the foot of the perpendicular drawn from (0, 0) to the line joining $({a\ cos}{\alpha,\ a\sin{\alpha)}}$ and $(a\cos{\beta,\ a\sin{\beta)}}\$ are
A) $\left( \frac{a}{2},\frac{b}{2} \right)$
B) $\left\lbrack \frac{a}{2}(\cos\alpha + \cos{\beta),\frac{a}{2}(\sin{\alpha + \sin{\beta)}}} \right\rbrack$
C) $\left\lbrack \cos{\frac{\alpha + \beta}{2},\ }\sin\frac{\alpha + \beta}{2} \right\rbrack$
D) $\left( 0,\frac{b}{2} \right)$
Explanation
(b)
$\because Slope\ of\ perpendicular = - \left\lbrack \frac{\cos\alpha - \cos\beta}{\sin\alpha - \sin\beta} \right\rbrack = \tan\frac{\alpha + \beta}{2}$
$\therefore$ Equation of perpendicular is
$y = \tan{\left( \frac{\alpha + \beta}{2} \right)x}\ \ \ \ \ \ldots(i)$
On solving the Eq. (i) with the line, we get
$\left\lbrack \frac{a}{2}\left( \cos\alpha + \cos\beta \right),\frac{a}{2}\left( \sin\alpha + \sin\beta \right) \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.