Question

The line $L$ has intercepts $a$ and $b$ on the co-ordinate axes. When keeping the origin fixed, the co-ordinate axes are rotated through a fixed angle, then the same line has intercepts $p$ and $q$ on the rotated axes. 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. $\frac{1}{a^2}+\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{q^q}$

Answer 1

Answer: (B)

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

Suppose the axes are rotated in the anti-clockwise direction through an angle $\alpha$. The equation of the line $L$ with respect to the old axes is given by $x/a+y/b=1$. To find the equation of $L$ with respect to the new axes, we replace $x$ by $xcos\alpha -ysin\alpha$ and $y$ by $xsin\alpha +ycos\alpha$, so that the equation of $L$ with respect to the new axes is $\frac{1}{a}(xcos\alpha -ysin\alpha )+\frac{1}{b}(xsin\alpha +ycos\alpha )=1$.
Since $p,q$ are the intercepts made by this line on the co=ordinate axes, we have on putting $(p,0)$ and then $(0,p)$
$1/p=\left(1/a\right)cos\alpha +\left(1/b\right)sin\alpha$
and $1/q=-\left(1/a\right)sin\alpha +\left(1/b\right)cos\alpha$
Eliminate $\alpha$
Squaring and adding, we get
$1/p^2+1/q^2=1/a^2+1/b^2$
$\frac{x}{a}+\frac{y}{b}=1$ transforms to $\frac{x}{p}+\frac{y}{q}=1$ after rotation of axes. Since origin and the lines are fixed and hence perpendicular from origin is same.
$\therefore \frac{1}{\sqrt{\left(1/a^2\right)+\left(1/b^2\right)}}=\frac{1}{\sqrt{\left(1/p^2\right)+\left(1/q^2\right)}}$
$\therefore \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$