Line L has intercepts a and b on the axes of coordinates, when the axesare rotated through a given angle, keeping the origin fixed. the same line L has intercepts p and q on the transformed axes, prove that $\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}=\frac{1}{{\mathrm{p}}^2}+\frac{1}{{\mathrm{q}}^2}$

Equation of L in the original system is $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1 ...\left(1\right)$
Equation of L after rotating the axes is $\frac{\mathrm{x}}{\mathrm{p}}+\frac{\mathrm{y}}{\mathrm{q}}=1 ...\left(2\right)$
The length of the perpendicular from the origin to the line in the both the system are the same
$\Rightarrow \frac{1}{\sqrt{\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}}}=\frac{1}{\sqrt{\frac{1}{{\mathrm{p}}^2}+\frac{1}{{\mathrm{q}}^2}}}\Rightarrow \frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}=\frac{1}{{\mathrm{p}}^2}+\frac{1}{{\mathrm{q}}^2}$