Question

Heyo! ʕ´•ᴥ•`ʔ

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 1/p² = 1/a² + 1/b².

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Answer 1

The equation of the line having intercepts on the x-axis and y-axis as a and b respectively, is

$\frac{x}{a} + \frac{y}{b} = 1$

That is, bx+ay−ab=0

The length of the perpendicular from origin onto this line is given by

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

Hope it helps!

Answer 2

Hope it will helps......❣️❣️

❤Sweetheart❤

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