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Question
If p be the length of the perpendicular from the origin on the line x/a + y/b = 1, then
Options
p2 = a2 + b2
\[p^2 = \frac{1}{a^2} + \frac{1}{b^2}\]
\[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\]
none of these
Solution
\[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\]
It is given that p is the length of the perpendicular from the origin on the line \[\frac{x}{a} + \frac{y}{b} = 1\]
\[\frac{1}{a}x + \frac{1}{b}y - 1 = 0\]
\[ \therefore p = \left| \frac{0 + 0 - 1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \right|\]
Squaring both sides,
\[ \Rightarrow \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\]
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