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AIEEE 2011

JEE Main 2011 Mathematics question on Straight Lines and Pair of Lines

A straight line LL through the point (3,−2)(3, -2) is inclined at an angle 60°60° to the line 3x+y=1\sqrt{3}x + y = 1. If LL also intersects the X-axis, then the equation of LL is
  1. A.
    y+3x+2−33=0y + \sqrt{3}x + 2 - 3\sqrt{3} = 0
  2. B.
    y−3x+2+33=0y - \sqrt{3}x + 2 + 3\sqrt{3} = 0
  3. C.
    3y−x+3+23=0\sqrt{3}y - x + 3 + 2\sqrt{3} = 0
  4. D.
    3y+x−3+23=0\sqrt{3}y + x - 3 + 2\sqrt{3} = 0
Show answer and solution

Correct option: B

Solution

The line √3x + y = 1 has slope m₁ = -√3, which corresponds to angle 120°. A line inclined at 60° to this line has direction angle either 120° + 60° = 180° (slope 0, horizontal — doesn't work since horizontal through (3,-2) is y = -2, which intersects x-axis only at infinity... actually it meets x-axis nowhere unless y=0, so reject) or 120° - 60° = 60°, giving slope tan(60°) = √3.

Using slope m = √3 through (3, -2):
y - (-2) = √3(x - 3)
y + 2 = √3x - 3√3
y - √3x + 2 + 3√3 = 0

This line has slope √3 > 0, so it rises and will intersect the x-axis (set y = 0: x = (2 + 3√3)/√3 = 2/√3 + 3, a real finite value). This matches option 2.

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