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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 1/p² = 1/a² + 1/b².
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The equation of the line having intercepts on the x-axis and y-axis as a and b respectively, is
That is,bx+ay-ab=0
The length of the perpendicular from origin into this line is given by
I hope it's help UH.....:)
Answered by
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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
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