# Line 'L' has interceps a and b on the coordinate axis. When the axis are rotated through a given angle, keeping the origin fixed, the same line 'L' has intercepts p and q on the new axis, thenA) a2 + b2 = p2 + q2B) a2 + p2 = b2 + q2C) 1/a2 + 1/b2 = 1/p2 + 1/q2D) 1/a2 + 1/p2 = 1/b2 + 1/q2

Pratham Ashish
17 Points
14 years ago

hi,

in the first coordinate sys,

x/a + y/b = 1     ..............................1

in other

x'/ p + y'/q  =1

where

\begin{align} x'&=x\cos\theta+y\sin\theta\\ y'&=-x\sin\theta+y\cos\theta. \end{align}

so,  x {  cos θ/p - sinθ /q }   + y { sinθ /p  + cosθ/ q }  =1

comparing with (1).....

{  cos θ/p - sinθ /q }/ 1/a   = { sinθ /p  + cosθ/ q }  / 1/b = 1/1

cos θ/p - sinθ /q   = 1/a  ...............2

sinθ /p  + cosθ/ q  = 1/b ........................3

1/p^2 +  1/q^2  = 1/a^2 + 1/b^2

14 Points
6 years ago
H
14 Points
6 years ago
Here the length of perpendicular from origin to the lines would be equal hence on equating that distance we get
√(1/p^2 +1/q^2)=√(1/a^2+1/b^2)
On squaring both sides we get the answer
4 years ago
Dear student,

Here the length of perpendicular from origin to the lines would be equal hence on equating that distance we get
√(1/p^2 +1/q^2)=√(1/a^2+1/b^2)
On squaring both sides we get the answer
Hence the correct option will be C

Thanks and regards,
Kushagra