Question

The line $L$ has intercepts $a$ and $b$ on the coordinate axes. When keeping the origin fixed, the coordinate axes are rotated through a fixed angle, then the same line has intercepts $p$ and $q$ on the rotated axes. Then prove that
$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$

Answer 1

Suppose the axes are rotated in the anti-clockwise direction through an angle $\alpha$. The equation of the line $L$ with respect to the old axes is given by $x/a+y/b=1$. To find the equation of $L$ with respect to the new axes, we replace $x$ by $x\cos \alpha -y\sin \alpha$ and $y$ by $x\sin \alpha +y\cos \alpha$, so that the equation of $L$ with respect to the new axes is
$\frac{1}{a}(x\cos \alpha -y\sin \alpha )+\frac{1}{b}(x\sin \alpha +y\cos \alpha )=1$
Since $p,q$ are the intercepts made by this line on the coordinate axes, we have on putting $(p,0)$ and then $(0,q)$
$1/p=\left(1/a\right)\cos \alpha +\left(1/b\right)\sin \alpha$
and $1/q=-\left(1/a\right)\sin \alpha +\left(1/b\right)\cos \alpha$
Eliminate $\alpha$
Squaring and adding we get
$1/p^2+1/q^2=1/a^2+1/b^2$