JEE Main 2019 Mathematics question on Straight Lines and Pair of Lines

The line x + y = 0 makes an angle of 135° with the positive x-axis. The perpendicular from the origin to line L makes an angle of 60° with this line, so the perpendicular's angle with x-axis is either 135° + 60° = 195° or 135° - 60° = 75°.

Since the line makes positive intercepts on both axes, the perpendicular from the origin must point into the first quadrant, so the angle is 75°.

The perpendicular from origin to L has direction 75°, so line L is perpendicular to this — it makes angle 75° - 90° = -15° (or 165°) with x-axis.

The foot of perpendicular from origin is at distance 4 in direction 75°:
foot = (4cos75°, 4sin75°)

cos75° = cos(45°+30°) = (√6-√2)/4, sin75° = (√6+√2)/4

So foot = (√6-√2, √6+√2).

The equation of line L passing through this foot with normal direction (cos75°, sin75°) is:
x·cos75° + y·sin75° = 4

Multiply through by 4: x(√6-√2) + y(√6+√2) = 16

Factor out √2: √2·x(√3-1) + √2·y(√3+1) = 16

Divide by √2: (√3-1)x + (√3+1)y = 16/√2 = 8√2

This is exactly option 4.