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

$x \: \: + \: y \: \: = 1\: \: \\ a \: \: \: \: \: \: \: \: \: \: b$

That is,bx+ay-ab=0

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

$p = |0 + 0 - ab| = ab\\ \sqrt{a2 \: + \: b2 \: } \: \: \: \sqrt{a2 + b2}$

$p2 = a2 \: b2 \\ \: \: \: a2 + b2$

$1 = 1 + 1\\ p2 \: \: \: \: b2 \: \: a2$

I hope it's help UH.....:)

Answer 2

Answer:

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

\begin{gathered}x \: \: + \: y \: \: = 1\: \: \\ a \: \: \: \: \: \: \: \: \: \: b\end{gathered}

x+y=1

ab

That is,bx+ay-ab=0

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

\begin{gathered}p = |0 + 0 - ab| = ab\\ \sqrt{a2 \: + \: b2 \: } \: \: \: \sqrt{a2 + b2} \end{gathered}

p=∣0+0−ab∣=ab

a2+b2

a2+b2

\begin{gathered}p2 = a2 \: b2 \\ \: \: \: a2 + b2\end{gathered}

p2=a2b2

a2+b2

\begin{gathered}1 = 1 + 1\\ p2 \: \: \: \: b2 \: \: a2\end{gathered}

1=1+1

p2b2a2

I hope it's help