Question

If $p$ is the length of the perpendicular from origin to the line $\frac{x}{a}+\frac{y}{b}=1$, then the correct relation between $a,b$ and $p$ is

• A. $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$
• B. $\frac{1}{p^2}=\frac{1}{a^2}-\frac{1}{b^2}$
• C. $\frac{1}{p^2}=\frac{a^2{b}^2}{a^2+b^2}$
• D. $\frac{1}{p^2}=-\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

The correct option is A $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Given: $\frac{x}{a}+\frac{y}{b}=1$
$\Rightarrow bx+ay=ab$
Dividing with
$\sqrt{A^2+B^2}=\sqrt{a^2+b^2}$
$\frac{bx}{\sqrt{a^2+b^2}}+\frac{ay}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{a^2+b^2}}$
So,
$|p|=\frac{ab}{\sqrt{a^2+b^2}}$
$\Rightarrow p^2=\frac{a^2{b}^2}{a^2+b^2}\Rightarrow \frac{1}{p^2}=\frac{a^2+b^2}{a^2{b}^2}\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$