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(A) The vertices of the triangle are the intersection points of the given lines.Solving the pairs of equations:$1$) $x + y = 4$ and $3x + y = 4$: Subt

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Isosceles

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(A) The vertices of the triangle are the intersection points of the given lines.Solving the pairs of equations:$1$) $x + y = 4$ and $3x + y = 4$: Subtracting gives $2x = 0 \implies x = 0, y = 4$. Vertex $A(0, 4)$.$2$) $3x + y = 4$ and $x + 3y = 4$: Solving gives $x = 1, y = 1$. Vertex $B(1, 1)$.$3$) $x + y = 4$ and $x + 3y = 4$: Subtracting gives $2y = 0 \implies y = 0, x = 4$. Vertex $C(4, 0)$.Now,calculate the lengths of the sides:$AB = \sqrt{(1-0)^2 + (1-4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{10}$$BC = \sqrt{(4-1)^2 + (0-1)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{10}$$AC = \sqrt{(4-0)^2 + (0-4)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}$Since $AB = BC$,the triangle is isosceles.

The triangle formed by the lines $x + y - 4 = 0,$ $3x + y = 4,$ and $x + 3y = 4$ is

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