the straight lines x+y=0, 3x+y-4=0 and x+3y-4=0 which is

1. isosless
2. equilateral
3. right angle
4. none

Since none of the lines are perpendicular therefore it is not a right angled triangle.
now find the intersection points of x+y=0, 3x+y-4=0, x+3y-4=0, we get
(2,-2),(-2,2) & (1,1)
let A(2,-2) B(-2,2), C(1,1)
By distance formula we get AC=AB but not equal to AB therefore it is a isoscels triangle.