Question

The line L has intercepts a and b on the coordinate axes. When keeping the origin fixed, the coordinate axes are rotated through a fixed angle, then the same line has intercepts p and q on the rotated axes. Then
(I.I.T. 1990)

• 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. $a^2+b^2=p^2+q^2$
• 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 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 $\frac{x}{a}+\frac{y}{b}=1$ To find the equation of L with respect to the new axes, we replace x by (x cos $\alpha$ - y sin $\alpha$) and y by (x sin $\alpha$ + y cos $\alpha$) 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, q)
$\frac{1}{p}=\frac{1}{a}cos\alpha +\frac{1}{b}sin\alpha$
and $\frac{1}{q}=-\frac{1}{a}sin\alpha +\frac{1}{b}cos\alpha$

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

Alternate solution:
$\frac{x}{a}+\frac{y}{b}=1$ transforms to $\frac{x}{p}+\frac{x}{q}=1$ after rotation of axes. Since origin and the line are fixed only its equation is changed and hence perpendicular from origin is same
$\therefore \frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{1}{\sqrt{\frac{1}{p^2}+\frac{1}{q^2}}}$
$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$