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JEE Main20207 Jan 2020Evening ShiftMathematicsStraight LinesActual

The locus of the mid-points of the perpendiculars drawn from points on the line x = 2 y , to the line x = y , is.

Options

  1. A2 x - 3 y = 0
  2. B5 x - 7 y = 0
  3. C3 x - 2 y = 0
  4. D7 x - 5 y = 0

Correct answer

B. 5 x - 7 y = 0

Step-by-step solution

Slope of P Q = k - α h - 2 α = - 1 k - α = - h + 2 α ⇒ α = h + k 3     … 1 Also, 2 h = 2 α + β and 2 k = α + β 2 h = α + 2 k ⇒ α = 2 h - 2 k     … 2 From 1   &   2 , h + k 3 = 2 h - k So, locus is 6 x - 6 y = x + y ⇒ 5 x = 7 y .

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