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JEE Main 1990 Mathematics question on Straight Lines and Pair of Lines

Line LL has intercepts aa and bb on the coordinate axes. When, the axes are rotated through a given angle, keeping the origin fixed, the same line LL has intercepts pp and qq, then
  1. A.
    a2+b2=p2+q2a^2 + b^2 = p^2 + q^2
  2. B.
    1a2+1b2=1p2+1q2\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2} + \frac{1}{q^2}
  3. C.
    a2+p2=b2+q2a^2 + p^2 = b^2 + q^2
  4. D.
    1a2+1p2=1b2+1q2\frac{1}{a^2} + \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{q^2}
Show answer and solution

Correct option: B

Solution

The equation of line L with intercepts a and b is x/a + y/b = 1, so its distance from the origin is 1/√(1/a² + 1/b²).

After rotation, the same line has intercepts p and q, giving distance from origin = 1/√(1/p² + 1/q²).

Since rotation keeps the origin fixed and doesn't change the line itself, the perpendicular distance from the origin to the line is invariant. Therefore 1/√(1/a² + 1/b²) = 1/√(1/p² + 1/q²), which gives 1/a² + 1/b² = 1/p² + 1/q².

[1990]

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