A line L makes intercepts a and b on the coor | vedclass

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(C) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.The perpendicular distance $d$ from the origi

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$\frac{1}{p^2}+\frac{1}{q^2}$

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(C) The equation of the line $L$ with intercepts $a$ and $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.The perpendicular distance $d$ from the origin $(0, 0)$ to this line is given by $d = \frac{|-1|}{\sqrt{(\frac{1}{a})^2 + (\frac{1}{b})^2}} = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.Squaring both sides,we get $\frac{1}{d^2} = \frac{1}{a^2} + \frac{1}{b^2}$.When the axes are rotated,the origin remains fixed,so the perpendicular distance $d$ from the origin to the line $L$ remains invariant.In the new coordinate system,the line $L$ makes intercepts $p$ and $q$,so its equation is $\frac{x}{p} + \frac{y}{q} = 1$.The perpendicular distance $d$ from the origin to this new line is $d = \frac{1}{\sqrt{\frac{1}{p^2} + \frac{1}{q^2}}}$,which implies $\frac{1}{d^2} = \frac{1}{p^2} + \frac{1}{q^2}$.Since $d$ is the same in both cases,we have $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}$.

$A$ line $L$ makes intercepts $a$ and $b$ on the coordinate axes. The axes are rotated through an angle $\theta$ in the positive direction,keeping the origin fixed. If the line $L$ makes intercepts $p$ and $q$ on the new coordinate axes,then $\frac{1}{a^2}+\frac{1}{b^2}=$

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