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# 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

12 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

3 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 Kushagra Madhukar
one year 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