Question

If length of perpendicular from origin to line $\frac{x}{a}+\frac{y}{b}=1$ is p, then prove that
$\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$ ?

Answer 1

use intercept form of line : $\frac{x}{a}+\frac{y}{b}=1$ and then find perpendicular distance of point from line .

equation of line in intercept form is

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

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

now, use formula

distance of point $(x1,y1)$ from line : $ax+by+c=0$ is $\frac{|ax1+by1+c|}{\sqrt{(a²+b²)}}$

so, it's distance from origin is

$P=\frac{|0+0-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$

$P=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$

$\frac{1}{P}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$

take square both sides,

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

hence proved.