Question

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

then which of the following is true?

• A. $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{p^2}=0$
• B. $a^2+b^2-p^2=0$
• C. $\frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{p^2}$
• D. $\frac{1}{a^2}+\frac{1}{b^2}-\frac{1}{p^2}=0$

Answer 1

Answer: (D)

$\frac{1}{a^2}+\frac{1}{b^2}-\frac{1}{p^2}=0$

Given line

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

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

p is the length of perpendicular from origin to given line

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

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

On squaring both sides

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

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

$\frac{1}{a^2}+\frac{1}{b^2}-\frac{1}{p^2}=0$