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JEE MainMathematicsStraight Lines

Let A be the point of intersection of the lines x + y = 6 and 2x - y = 6 . A variable line passes through A . The locus of the foot of the perpendicular drawn from the origin O(0,0) to this variable line forms a curve S . The perpendicular distance from the center of the curve S to the line 3x + 4y + 10 = 0 is

Options

  1. A4
  2. B6
  3. C10
  4. D0

Correct answer

A. 4

Step-by-step solution

Solving the given lines x + y = 6 and 2x - y = 6 , we add them to get 3x = 12 x = 4 . Substituting x = 4 into x + y = 6 , we get y = 2 . Thus, the point A is (4, 2) . Let the foot of the perpendicular from the origin O(0,0) to the variable line passing through A(4,2) be P(h, k) . Since OP is perpendicular to the line AP , the product of their slopes is -1 . Slope of OP = k h and Slope of AP = k - 2 h - 4 . ( k h ) ( k - 2 h - 4 ) = -1 k(k - 2) = -h(h - 4) h^2 - 4h + k^2 - 2k = 0 Replacing (h, k) with (x, y) , the l

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