Question

If p be the length of the perpendicular from the origin on the line $\frac{x}{a}+\frac{y}{b}=1,$ then

• A. $p^2=a^2+b^2$
• 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

The correct option is 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

$\frac{x}{a}+\frac{y}{b}=1$

$\frac{1}{a}x+\frac{1}{b}y-1=0$

$\therefore p=\left|\frac{0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right|$

Squaring both sides,

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