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

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

$\Rightarrow b x + a y = a b \Rightarrow b x + a y - a b = 0$

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

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

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

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

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