The straight lines $x+y-4=0, 3x+y-4 =0, x+3y-4 =0$ forms a triangle which is

Straight lines $x+y -4 = 0, 3x+y-4=0, x+3y-4=0$ form a triangle.

Point of interaction of line $x+y-4=0$ and $3x + y - 4= 0$ can be calculated by solving and we will get point of interaction $A = (2, -2)$

Point of interaction of line $3x+y-4=0$ and $x +3 y - 4= 0$ can be calculated by solving and we will get point of interaction $B = (1, 1)$

Point of interaction of line $x+3y-4=0$ and $x + y - 4= 0$ can be calculated by solving and we will get point of interaction $c = (-2, 2)$

Hence we get a triangle $\triangle ABC$

Using distance formula, we can calculate length of sides $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ between points $(x_1, y_1)$ and$(x_2,y_2)$

length of AB $= \sqrt{(1-2)^2 + (1+2)^2} \\ = \sqrt{1+9} \\ = \sqrt{10}$units

length of BC $= \sqrt{(-2-1)^2 + (2-1)^2} \\ = \sqrt{9+1} \\ = \sqrt{10}$units

length of CA $= \sqrt{(-2-2)^2 + (2+2)^2} \\ = \sqrt{16+16} \\ = \sqrt{32}$units

As we can observe, AB=BC, the triangle is isosceles.