If the straight lines $$x + 3y = 4, 3x + y = 4$$ and $$x + y = 0$$ form a triangle, then the triangle is

The three sides of the triangle are the pair-wise intersections of the given lines.

Vertex A: intersection of $$x+3y=4$$ and $$3x+y=4$$.
Solve simultaneously: $$x+3y=4 \quad -(1)$$ $$3x+y=4 \quad -(2)$$ Multiply (1) by 3 and subtract (2): $$8y=8 \Rightarrow y=1$$. Insert in (2): $$3x+1=4 \Rightarrow x=1$$.
Hence $$A(1,\,1)$$.

Vertex B: intersection of $$3x+y=4$$ and $$x+y=0$$. Subtract: $$2x=4 \Rightarrow x=2$$, then $$y=-x=-2$$. So $$B(2,\,-2)$$.

Vertex C: intersection of $$x+3y=4$$ and $$x+y=0$$. Subtract: $$2y=4 \Rightarrow y=2$$, then $$x=-y=-2$$. Thus $$C(-2,\,2)$$.

Now compute the side lengths using the distance formula $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$.

$$AB=\sqrt{(2-1)^2+(-2-1)^2}=\sqrt{1^2+(-3)^2}=\sqrt{10}$$

$$AC=\sqrt{(-2-1)^2+(2-1)^2}=\sqrt{(-3)^2+1^2}=\sqrt{10}$$

$$BC=\sqrt{(-2-2)^2+(2+2)^2}=\sqrt{(-4)^2+4^2}=\sqrt{16+16}=4\sqrt{2}$$

Two sides are equal: $$AB=AC=\sqrt{10}$$, while $$BC=4\sqrt{2}$$ is different. Hence the triangle is isosceles.

To check for a right angle, note that the slopes of $$AB$$ and $$AC$$ are $$-3$$ and $$-\frac13$$, whose product is $$1 \neq -1$$. Therefore no angle is $$90^{\circ}$$, so the triangle is not right-angled.

Thus the triangle is isosceles.

Option C which is: isosceles