Question

If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

The given line is $\frac{x}{a}+\frac{y}{b}=1$.

$\Rightarrow bx+ay=ab\Rightarrow bx+ay-ab=0$

Now p is the length of perpendicular from origin to bx + ay - ab = 0

$\therefore p=\left|\frac{b\times 0+a\times 0-ab}{\sqrt{b^2+a^2}}\right|=\frac{ab}{\sqrt{b^2+a^2}}$

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

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