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Question: If the length of the perpendicular drawn from origin to the line whose intercepts on the axes are *a...

If the length of the perpendicular drawn from origin to the line whose intercepts on the axes are a and b be p, then

A

a2+b2=p2a^{2} + b^{2} = p^{2}

B

a2+b2=1p2a^{2} + b^{2} = \frac{1}{p^{2}}

C

1a2+1b2=2p2\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{2}{p^{2}}

D

1a2+1b2=1p2\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}}

Answer

1a2+1b2=1p2\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}}

Explanation

Solution

Equation of line is xa+yb=1\frac{x}{a} + \frac{y}{b} = 1bx+ayab=0bx + ay - ab = 0

Perpendicular distance from origin to given line is

p=aba2+b2p = \left| \frac{- ab}{\sqrt{a^{2} + b^{2}}} \right|a2+b2ab=1p\frac{\sqrt{a^{2} + b^{2}}}{ab} = \frac{1}{p}a2+b2a2b2=1p2\frac{a^{2} + b^{2}}{a^{2}b^{2}} = \frac{1}{p^{2}}

1a2+1b2=1p2\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}}