If the lines x + 3y = 4,3x + y = 4,and x + y | vedclass

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(C) The given lines are:$L_1: x + y = 0$$L_2: 3x + y = 4$$L_3: x + 3y = 4$Solving $L_1$ and $L_2$: $x + y = 0 \Rightarrow y = -x$. Substituting into $

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Isosceles

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(C) The given lines are:$L_1: x + y = 0$$L_2: 3x + y = 4$$L_3: x + 3y = 4$Solving $L_1$ and $L_2$: $x + y = 0 \Rightarrow y = -x$. Substituting into $L_2$: $3x - x = 4$ $\Rightarrow 2x = 4$ $\Rightarrow x = 2, y = -2$. Vertex $A = (2, -2)$.Solving $L_2$ and $L_3$: $y = 4 - 3x$. Substituting into $L_3$: $x + 3(4 - 3x) = 4$ $\Rightarrow x + 12 - 9x = 4$ $\Rightarrow -8x = -8$ $\Rightarrow x = 1, y = 1$. Vertex $B = (1, 1)$.Solving $L_1$ and $L_3$: $y = -x$. Substituting into $L_3$: $x + 3(-x) = 4$ $\Rightarrow -2x = 4$ $\Rightarrow x = -2, y = 2$. Vertex $C = (-2, 2)$.Calculating side lengths:$AB = \sqrt{(1-2)^2 + (1 - (-2))^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$$BC = \sqrt{(-2-1)^2 + (2-1)^2} = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$$AC = \sqrt{(-2-2)^2 + (2 - (-2))^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32}$Since $AB = BC = \sqrt{10}$,the triangle is isosceles.

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

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