Question

The line $L$ has intercepts $a$ and $b$ on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If $p$ and $q$ are the intercepts of the line $L$ on the new axes, then $\frac{1}{a^2}-\frac{1}{p^2}+\frac{1}{b^2}-\frac{1}{q^2}$equal to

• A. $-1$
• B. $0$
• C. $1$
• D. none of these

Answer 1

The correct option is B $0$

Equation of the line $L$ in the two coordinate systems is $\frac{x}{a}+\frac{y}{b}=1,\frac{X}{p}+\frac{Y}{q}=1$ where $(X,Y)$ are the new coordinates of a point $(x,y)$ when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed.
$\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}$ $\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$ or $\frac{1}{a^2}-\frac{1}{p^2}+\frac{1}{b^2}-\frac{1}{q^2}=0$