show that the lines x2-4xy+y2=0 and x+y=3 form an equilateral triangle also find area of the triangle

Given pair of lines are x² + 4xy + y² = 0

⇒ (y/x) ² + 4 y/x + 1 = 0

⇒ y/x = -4±2√3/2 = -2±√3,

∴  The lines y = (-2 + √3) x and y = (-2 - √3) x and x - y = 4 forms an equilateral triangle

Clearly the pair of lines x² + 4xy +y²  = 0  intersect at origin,

The perpendicular distance form origin to x - y = 4 is the height of the

h = 2 √ 2

∵ Area of triangle = h²/√3 = 8/√3