show that the lines x^-4xy+y^=0 and x+y=1 form an equilateral triangle and find its area

Answer 1

If that's x^2 - 4xy + y^2 = 0, that means that

y = (2+√3)x and y = (2-√3)x

Find where those lines intersect y = -x + 1

A: (2+√3)x = 1-x
x = 1/(3+√3)
y = 1-x = (2+√3)/(3+√3)

B:(2-√3)x = 1-x
x = 1/(3-√3)
y = 1 - x = (2-√3)/(3-√3)

After a little algebra, we find that

|OA|² = 2/3
|OB|² = 2/3

|AB|^2 = √6/3

The altitude = 1/√2, so the area of OAB = 1/2 * 1/√2 * √6/3 = √3/6

Answer 2

That's |AB| = √6/3