Question

The line $L$ has intercepts $a$ and $b$ on the co-ordinate axes keeping the origin fixed, the co-ordinate axes are related through a fixed angle. If the same line has intercepts c and d then

• A. $\frac{1}{a^2}+\frac{1}{c^2}=\frac{1}{b^2+d^2}$
• B. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}+\frac{1}{d^2}$
• C. $a^2+c^2=b^2+d^2$
• D. $a^2+b^2=c^2+d^2$

Answer 1

The correct option is B $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}+\frac{1}{d^2}$

Suppose we state he coordinate axis in the anti-clockwise direction through an angle $\alpha$.
The equation of the line $\alpha$ with respect to old axes is $\frac{x}{a}+\frac{y}{b}=1$
In this equation replacing $x$ by $x\cos \alpha -y\sin \alpha$
The equation of the line with respect to new axes is
$\frac{x\cos \alpha -y\sin \alpha }{a}+\frac{x\sin \alpha +y\cos \alpha }{b}=1$
$\Rightarrow x(\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b})+y(\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a})=1$ ...(1)
The intercept mode by (1) on the co-ordinate axes are given as $c$ and $d$.
Therefore, $\frac{1}{c}=\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}$ and $\frac{1}{d}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}$
Squaring and adding, we get $\frac{1}{c^2}+\frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2}$