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

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Solution

$G i v e n i n t e r c e p t s a r e$ $a$ and $b$ . $H e n c e e q u a t i o n o f l i n e i s$ $\frac{x}{a} + \frac{y}{b} = 1$ .

$P e r p e n d i c u l a r d i s t a n c e o r i g i n i s$ $(x_{1}, y_{1})$ from line $a x + b y + c = 0$ is
$\Rightarrow$ $∣ ∣ ∣ \frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} ∣ ∣ ∣$

$\Rightarrow$ $p = \frac{∣ ∣ ∣ \frac{0}{a} + \frac{0}{b} - 1 ∣ ∣ ∣}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$
$\Rightarrow$ $p = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} \Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

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If p is the length of perpendicular from the origin to the line whose intercepts on the axis are a and b, then show that 1p2=1a2+1b2.

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