Question

If p be the length of the perpendicular from the origin on the line x/a + y/b = 1, then

विकल्प

•  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

Answer 1

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

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, 

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