Q. If $P$ is the length of the perpendicular from the origin on the line whose intercepts on the axes are $a$ and $b$, then
A) $p^{2} = a^{2} + b^{2}$
B) $p^{2} = a^{2} - b^{2}$
C) $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$
D) $\frac{1}{p^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}$
Explanation
(c)
$Here\ the\ equation\ of\ AB\ is\frac{x}{a} + \frac{y}{b} = 1$
From the figure, $OP\bot AB,$
[image:10_image22.emf]
$\therefore OP = \left| \frac{0\left( \frac{1}{a} \right) + 0\left( \frac{1}{b} \right) - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} \right|$
$\Rightarrow p = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$
$\Rightarrow \ p^{2} = \frac{1}{\frac{1}{a^{2}} + \frac{1}{b^{2}}}\ \lbrack squaring\ both\ sides\rbrack$
$\Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.