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)
1a2+1b2=1p2+1q2
Suppose the axes are rotated in the anti- clockwise direction through an angle α. The equation of the line L with respect to the old axes is given by xa+yb=1 To find the equation of L with respect to the new axes, we replace x by (x cos α - y sin α) and y by (x sin α + y cos α) so that the equation of L with respect to the new axes is 1a(xcosα−ysinα)+1b(xsinα+ycosα)=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)
1p=1acosα+1bsinα
and 1q=−1asinα+1bcosα
Eliminate α. Squaring and adding, we get
1p2+1q2=1a2+1b2
Alternate solution:
xa+yb=1 transforms to xp+xq=1 after rotation of axes. Since origin and the line are fixed only its equation is changed and hence perpendicular from origin is same
∴ 1√1a2+1b2=1√1p2+1q2
1a2+1b2=1p2+1q2