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(C) The slopes of the given lines are $m_1 = -1$,$m_2 = -3$,and $m_3 = -\frac{1}{3}$.Let the vertices be $A$,$B$,and $C$. The intersection points are:

Vedclass · Accepted Answer

Isosceles

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(C) The slopes of the given lines are $m_1 = -1$,$m_2 = -3$,and $m_3 = -\frac{1}{3}$.Let the vertices be $A$,$B$,and $C$. The intersection points are:$1$. $x + y = 0$ and $3x + y = 4$: $2x = 4 \implies x = 2, y = -2$. Vertex $P(2, -2)$.$2$. $3x + y = 4$ and $x + 3y = 4$: Solving gives $x = 1, y = 1$. Vertex $Q(1, 1)$.$3$. $x + y = 0$ and $x + 3y = 4$: $2y = 4 \implies y = 2, x = -2$. Vertex $R(-2, 2)$.Calculate the lengths of the sides:$PQ = \sqrt{(2-1)^2 + (-2-1)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.$QR = \sqrt{(1 - (-2))^2 + (1-2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.$PR = \sqrt{(2 - (-2))^2 + (-2-2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32}$.Since $PQ = QR = \sqrt{10}$,two sides are equal.Therefore,the triangle is an isosceles triangle.

What type of triangle is formed by the lines $x + y = 0$,$3x + y = 4$,and $x + 3y = 4$?

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