Question

A line has intercepts a, b on the coordinate axes. If the axes are rotated about the origin through an angle $\alpha$ then the line has intercepts p,q on the new position of the axes respectively. Then

• A. $\frac{1}{p^2}+\frac{1}{q^2}=\frac{1}{a^2}+\frac{1}{b^2}$
• B. $\frac{1}{p^2}-\frac{1}{q^2}=\frac{1}{a^2}-\frac{1}{b^2}$
• C. $\frac{1}{p^2}+\frac{1}{a^2}=\frac{1}{q^2}+\frac{1}{b^2}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{p^2}+\frac{1}{q^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Since the line has intecepts a and b on the coordinate axes, therefore its equation is
$\frac{x}{a}+\frac{y}{b}=1$
When the axes are rotated, its equation with respect to the new axes and the same origin will become
$\frac{x}{p}+\frac{y}{q}=1$
In both the cases, the length of the perpendicular from the origin to the line will be same
Therefore
$\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$