Question

If $p$ is the perpendicular from origin to the line $\frac{x}{a}+\frac{y}{b}=1$, then

• 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. $\frac{1}{p^2}=-\frac{1}{a^2}-\frac{1}{b^2}$
• 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 line is
$\frac{x}{a}+\frac{y}{b}=1$
Thus, the length of perpendicular from origin to the line $\frac{x}{a}+\frac{y}{b}=1$ is
$p=\frac{b\times 0+a\times 0-ab}{\sqrt{b^2+a^2}}$
According to question,
$\Rightarrow \frac{-ab}{\sqrt{b^2+a^2}}=p$
$\Rightarrow \frac{(ab{)}^2}{b^2+a^2}=p^2$
$\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.