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NTA Abhyas JEE Main2020MathematicsStraight LinesPractice

Consider the family of lines 5 x + 3 y - 2 + λ 3 x - y - 4 = 0 and x - y + 1 + μ 2 x - y - 2 = 0 . The equation of a straight line that belongs to both the families is

Options

  1. A5 x - 2 y - 7 = 0
  2. B3 x + y - 2 = 0
  3. C5 x + 2 y - 3 = 0
  4. D2 x + y - 1 = 0

Correct answer

A. 5 x - 2 y - 7 = 0

Step-by-step solution

All lines of the family 5 x + 3 y - 2 = 0 and 3 x - y - 4 = 0 are concurrent. The point of concurrency is 1 , - 1 Similarly, the point of concurrency of lines of the other family is 3,4 ⇒ The line of both families is the line passing through 1 , - 1 & 3,4 ⇒ Equation of the required line is 5 x - 2 y - 7 = 0

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