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

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Solution

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

$\frac{x}{a} + \frac{y}{b} = 1, i . e . \frac{x}{a} + \frac{y}{b} - 1 = 0$

p = $∣ ∣ ∣ ∣ \begin{matrix} \frac{\frac{1}{a} \times 0 + \frac{1}{b} \times 0 - 1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} \end{matrix} ∣ ∣ ∣ ∣$

= $\frac{| - 1 |}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}}$

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

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

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

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