Geo2.33

If the perpendicular distance of the straight line $\frac{x}{a}+\frac{y}{b}=1\,$ is p, prove that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\,$ .

Wrting the given equation in general form,i tlooks like

$bx+ay=ab\,$

The perpendicular distance from the origin to this equation is

$|\frac{b\cdot 0+a\cdot 0-ab}{\sqrt{a^2+b^2}}|=p\,$

$\frac{ab}{\sqrt{a^2+b^2}}=p\,$

Taking inverse on both sides and squaring,we get

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

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

Hence proved