Question

A line has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through an angle $\alpha$ in anticlockwise direction, keeping the origin fixed, the line makes equal intercepts on the coordinate axes. Then the value of $\cot \alpha$ is

• A. $\frac{a+b}{a-b}$
• B. $\frac{a-b}{a+b}$
• C. $a^2-b^2$
• D. $a^2+b^2$

Answer 1

The correct option is A $\frac{a+b}{a-b}$

Original equation of straight line is :
$\frac{x}{a}+\frac{y}{b}=1$
Its transformed equation is :
$\frac{x\cos \alpha -y\sin \alpha }{a}+\frac{x\sin \alpha +y\cos \alpha }{b}=1$
$\Rightarrow (\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b})x+(\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a})y=1$
Given this line has equal intercepts.
$\therefore \frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}$
$\Rightarrow b\cos \alpha +a\sin \alpha =a\cos \alpha -b\sin \alpha$
$\Rightarrow (a-b)\cos \alpha =(a+b)\sin \alpha$
$\Rightarrow \cot \alpha =\frac{a+b}{a-b}$