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

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Toolbox:
• Equation of a line whose $x$ and $y$ intercepts are $a$ and $b$ is $\large\frac{x}{a}$$+\large\frac{y}{b}$$=1$
• The perpendicular distance of a line $Ax+By+C=0$ from a point $(x_1, y_1)$ is given by $d=\large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
Equation of a line whose intercepts on the axes are $a$ and $b$ is $\large\frac{x}{a}$$+ \large\frac{y}{b}$$=1$
This can be written as
$bx+ay=ab$
or $bx+ay-ab=0$------------(1)
The perpendicular distance $d$ of a line $Ax+By+C=0$ from a point $(x_1, y_1)$ is $d=\large\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$------------(2)
Now comparing equation (1) and (2) we get,
$A=a, B=b$ and $c=-ab$
$\therefore$ perpendicular distance of the line passing through the point (0,0) is $p = \large\frac{|A(0)+B(0)-ab|}{\sqrt{a^2+b^2}}$
$\therefore p = \large\frac{|-ab|}{\sqrt{a^2+b^2}}$
Squaring on both sides we get,
$p^2=\large\frac{(ab)^2}{a^2+b^2}$
$\Rightarrow p^2(a^2+b^2)=a^2b^2$