The co-ordinate axes are rotated about the origin O in the counter-clockwise direction through an angle 60° If p and q are the intercepts made on the new axes by a straight line whose equation referred to original axes is $x+y=1\text{, then }\frac{1}{p^2}+\frac{1}{q^2}=$ 

If the axes are rotated about the origin in anticlockwise direction, the relations relating old coordinates (x, y) and new coordinates (X, Y) are

$\begin{array}{l}x=X\cos {60}^{\circ }-Y\sin {60}^{\circ }\text{ and }y=X\sin {60}^{\circ }+Y\cos {60}^{\circ }\\ \Rightarrow x=\frac{X-\sqrt{3}Y}{2}\text{ and }y=\frac{\sqrt{3}X+Y}{2}\end{array}$

The equation of the line$x + y = 1$with reference to new coordinate axes is

$\frac{X-\sqrt{3}Y}{2}+\frac{\sqrt{3}X-Y}{2}=1\Rightarrow (\sqrt{3}+1)X+(1-\sqrt{3})Y=2$

This intercepts lengths$p$ and $q$ on the new-coordinate axes

$\begin{array}{l}\therefore p=\frac{2}{\sqrt{3}}+1\text{ and }q=\frac{2}{1-\sqrt{3}}\\ \Rightarrow \frac{1}{p^2}+\frac{1}{q^2}=\frac{(\sqrt{3}+1{)}^2+(1-\sqrt{3}{)}^2}{4}=2\end{array}$