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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, then
A) a2 + b2 = p2 + q2
B) a2 + p2 = b2 + q2
C) 1/a2 + 1/b2 = 1/p2 + 1/q2
D) 1/a2 + 1/p2 = 1/b2 + 1/q2
10 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


10 years ago
							H

one year 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

one year ago
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