Question

A line $L$ has intercepts $a$ and $b$ on the co-ordinate axes. When the axes are rotated through an angle, keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$. Which of the following options is correct?

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

Answer 1

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

The equation of line $L$ with reference to the old axis is
$\frac{x}{a}+\frac{y}{b}=1$
$OM=$ Length of the perpendicular from $O(0,0)$ on it.
$\Rightarrow OM=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\cdots \left(i\right)$

Let the axes be rotated through an angle $\theta$. With reference to the new axes, the equation of the line $L$ is

$\frac{x}{p}+\frac{y}{q}=1$

$\therefore OM=$ Length of the perpendicular from $O(0,0)$ on it.

$\Rightarrow OM=\frac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}\cdots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

$\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$

This is the required relation.