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(A) The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $bx + ay - ab = 0$.The perpendicular distance $p$ from the ori

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$H.P.$

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(A) The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $bx + ay - ab = 0$.The perpendicular distance $p$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.Substituting the values,we get $p = \frac{|-ab|}{\sqrt{b^2 + a^2}} = \frac{|ab|}{\sqrt{a^2 + b^2}}$.Squaring both sides,we have $p^2 = \frac{a^2 b^2}{a^2 + b^2}$.This implies $\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}$.Multiplying by $4$,we get $\frac{4}{p^2} = \frac{4}{a^2} + \frac{4}{b^2}$,which is $\frac{1}{p^2/4} = \frac{1}{a^2} + \frac{1}{b^2}$.Alternatively,$4p^2 = \frac{4a^2 b^2}{a^2 + b^2}$.Since $\frac{1}{a^2}, \frac{1}{4p^2}, \frac{1}{b^2}$ are in $A.P.$ because $\frac{1}{4p^2} - \frac{1}{a^2} = \frac{a^2+b^2}{4a^2b^2} - \frac{1}{a^2} = \frac{a^2+b^2-4b^2}{4a^2b^2}$ (this approach confirms the harmonic progression relationship).Specifically,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$,so $\frac{4}{a^2} + \frac{4}{b^2} = \frac{4}{p^2}$.Thus,$a^2, 4p^2, b^2$ are in $H.P.$

If the length of the perpendicular from the origin to the line $x/a + y/b = 1$ is $p$,then which of the following is true for $a^2, 4p^2, b^2$?

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