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JEE Advanced2013MathematicsStraight LinesActual

For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than 2 2 .Then

Options

  1. Aa + b - c > 0
  2. Ba - b + c < 0
  3. Ca - b + c > 0
  4. Da + b - c < 0

Correct answer

A. a + b - c > 0

Step-by-step solution

a - b x + b - a y = 0 &#8658; x = y &#8658; Point of intersection - c a + b , - c a + b Now using distance formula between two points d = x 1 - x 2 2 + y 1 - y 2 2 &#8658; 1 + c a + b 2 + 1 + c a + b 2 &#60; 2 2 &#8658; 2 a + b + c a + b &#60; 2 2 &#8658; a + b + c a + b &#60; 2 &#8658;a + b + c &#60; 2 a + b &#8658; a + b - c &#62; 0

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