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

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(C) Let the equations of the lines be:$L_1: x + 3y = 4$$L_2: 3x + y = 4$$L_3: x + y = 0$Solving the equations to find the vertices:Intersection of $L_

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isosceles

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(C) Let the equations of the lines be:$L_1: x + 3y = 4$$L_2: 3x + y = 4$$L_3: x + y = 0$Solving the equations to find the vertices:Intersection of $L_1$ and $L_2$: $x + 3y = 4$ and $3x + y = 4 \implies x = 1, y = 1$. So,$B = (1, 1)$.Intersection of $L_1$ and $L_3$: $x + 3y = 4$ and $x + y = 0 \implies x = 2, y = -2$. So,$C = (2, -2)$.Intersection of $L_2$ and $L_3$: $3x + y = 4$ and $x + y = 0 \implies x = 2, y = -2$ (Wait,let's re-calculate).Correct vertices:$L_1 \cap L_2: x=1, y=1 \implies (1, 1)$$L_1 \cap L_3: x+3(-x)=4 \implies -2x=4 \implies x=-2, y=2 \implies (-2, 2)$$L_2 \cap L_3: 3x+(-x)=4 \implies 2x=4 \implies x=2, y=-2 \implies (2, -2)$Let $A = (-2, 2)$,$B = (1, 1)$,$C = (2, -2)$.Lengths of sides:$AB = \sqrt{(1 - (-2))^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$$BC = \sqrt{(2 - 1)^2 + (-2 - 1)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$$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 straight lines $x + 3y = 4$,$3x + y = 4$,and $x + y = 0$ form a triangle,then the triangle is

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