Question

If $p$ is the length of the perpendicular from the origin on the line whose intercepts on the axes $a$ 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. $\frac{1}{p^2}=\frac{1}{a^2}-\frac{1}{b^2}$

Answer 1

Answer: (C)

$\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

The equation of the line having intercepts on the $x$-axis and $y$-axis as $a$ and $b$ respectively, is $\frac{x}{a}+\frac{y}{b}=1$

i.e. $bx+ay-ab=0$

The length of the perpendicular from origin onto this line is given by $p=\frac{|0+0-ab|}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{a^2+b^2}}$

$\Rightarrow p^2=\frac{a^2{b}^2}{a^2+b^2}$

$\Rightarrow \frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2}$