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

Equation of line whose intercept on the axes are $a$ & $b$ is
$\frac{x}{a}+\frac{y}{b}=1$
$\Rightarrow \left(\frac{1}{a}\right)x+\left(\frac{1}{b}\right)y-1=0$
The perpendicular distance $\left(d\right)$ of a line $Ax+By+C=0$ from a point $(x_1,y_1)$ is given by $d=\frac{\left|\begin{array}{c}A{x}_1+B{y}_1+C\end{array}\right|}{\sqrt{A^2+B^2}}$
Here, $A=\frac{1}{a},B=\frac{1}{b}\&C=-1$
Given, distance from origin $(0,0)$ to the line $\frac{x}{a}+\frac{y}{b}=1$ is $p$
$\Rightarrow p=\frac{\left|\begin{array}{c}\left(0\right)\left(\frac{1}{a}\right)+\left(0\right)\left(\frac{1}{b}\right)-1\end{array}\right|}{\sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}}$
$\Rightarrow p=\frac{\left|\begin{array}{c}0+0-1\end{array}\right|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
$\Rightarrow p=\frac{\left|\begin{array}{c}-1\end{array}\right|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
$\Rightarrow p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
$\Rightarrow \frac{1}{p}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$
Squaring both sides
$\Rightarrow {\left(\frac{1}{p}\right)}^2={\left(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}\right)}^2$
$\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$
Hence proved.