Question

If p be the length of the perpendicular from the origin on the line x/a + y/b = 1, then
(a) p² = a² + b²

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

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

(d) none of these

Answer 1

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

It is given that p is the length of the perpendicular from the origin on the line $\frac{x}{a}+\frac{y}{b}=1$
$$\begin{gathered} \frac{1}{a}x+\frac{1}{b}y-1=0 \\ \therefore p=\left|\frac{0+0-1}{\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}{\displaystyle }}\right| \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides}, \\ \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} \end{gathered}$$