JEE MainMathematicsStraight Lines
Consider the family of lines x(1 + 2 ) + y(2 - ) = 4 + 3 , where is a real parameter. Let L be the line of this family which is at a maximum distance from the point A(3, 5) . If S is the area of the triangle formed by the line L and the coordinate axes, then the value of 12S is
Options
- A108
- B36
- C54
- D74
Correct answer
C. 54
Step-by-step solution
The given family of lines is x(1 + 2 ) + y(2 - ) = 4 + 3 . Rearranging the terms, we get: (x + 2y - 4) + (2x - y - 3) = 0 This represents a family of lines passing through the point of intersection of the lines: x + 2y - 4 = 0 2x - y - 3 = 0 Solving these two equations, we obtain the fixed point P(2, 1) . The line L belonging to this family which is at a maximum distance from the point A(3, 5) must be perpendicular to the line segment AP . The slope of AP is m_ AP = 5 - 1 3 - 2 = 4 . Since L is perpendicular to AP