Question

A line has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through an angle $\alpha$, keeping the origin fixed, the line makes equal intercepts on the coordinate axes.

Answer 1

equation of line in intercept form is $\frac{x}{a}+\frac{y}{b}=1$ ....(1)
Since, the axes is rotated by $\alpha$
Therefore, $x=X\cos \alpha -Y\sin \alpha$ and $y=X\sin \alpha +Y\cos \alpha$
Substituting in (1), we get
$\frac{X\cos \alpha -Y\sin \alpha }{a}+\frac{X\sin \alpha +Y\cos \alpha }{b}=1$
$\Rightarrow (\frac{\cos \alpha }{a}-\frac{\sin \alpha }{b})X+(\frac{\cos \alpha }{b}+\frac{\sin \alpha }{a})Y=1$
Since, the intercepts of the transformed co-ordinate systems are equal.
Therefore, $\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}$
$\Rightarrow \tan \alpha =\frac{a-b}{a+b}$
A) if $a=1,b=0$
Then $\tan \alpha =\frac{1-0}{1+0}=1\Rightarrow \alpha =\frac{\pi }{4}$
B) if $a=0,b=1$
Then $\tan \alpha =\frac{0-1}{0+1}=-1\Rightarrow \alpha =-\frac{\pi }{4}$
C) if $\frac{a}{b}=\frac{1+\sqrt{3}}{1-\sqrt{3}}$
Then $\tan \alpha =\sqrt{3}\Rightarrow \alpha =\frac{\pi }{3}$
B) if $a=2,b=0$
Then $\tan \alpha =\frac{2-0}{2+0}=1\Rightarrow \alpha =\frac{\pi }{4}$