Question

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

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

Answer 1

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

Let the equation of line be $\frac{x}{a}+\frac{y}{b}=1.$

When axes are rotated through an angle $\alpha$, the new coordinates $XY$ are related to old coordinates $xy$ as follows:

$x=X\cos \alpha -Y\sin \alpha$

$y=X\sin \alpha +Y\cos \alpha$

Substituting these values in the equation of line, we get

$\frac{X\cos \alpha -Y\sin \alpha }{a}+\frac{X\sin \alpha +Y\cos \alpha }{b}=1\Rightarrow X(\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b})+Y(\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a})=1$

As it makes equal intercepts in the new coordinate system, we get

$\Rightarrow \frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}$

$\Rightarrow \cos \alpha (b-a)=-\sin \alpha (a+b)$

$\Rightarrow \tan \alpha =\frac{a-b}{a+b}$