Question

Line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle keeping the origin fixed, the same line $L$ has intercepts $p$ and $q$.Then,

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

Suppose we rotate the coordinate axes in the anticlockwise direction through an angle.

The equation of the line $L$ with respect to old axes is

$\frac{x}{a}+\frac{y}{b}=1$

In this equation, replacing $x$ by $x\cos \alpha -y\sin \alpha$ and $y$ by $x\sin \alpha +y\cos \alpha$, the equation of the line with respect to new axes is

$\frac{x\cos \alpha -y\sin \alpha }{a}+\frac{x\sin \alpha +y\cos \alpha }{b}=1$

$x(\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b})+y(\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a})=1$ ...(1)

The intercepts made by (1) on the coordinate axes are given as $p$ and $q$.

Therefore,
$\frac{1}{p}=\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b}$

and $\frac{1}{q}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a}$

Squaring and adding, we get
$\frac{1}{p^2}+\frac{1}{q^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Hence, option B.