Step-by-step Solution:

Step 1: Find the Points of Intersection First, find the points where the lines intersect to determine the vertices of the triangle. 1. Intersection of \( x + y = 0 \) and \( 3x + y - 4 = 0 \): \[ \begin{cases} x + y = 0 \\ 3x + y = 4 \end{cases} \] Subtract the first equation from the second: \[ 2x = 4 \Rightarrow x = 2 \] Substitute \( x = 2 \) into \( x + y = 0 \): \[ 2 + y = 0 \Rightarrow y = -2 \] Intersection point: \( (2, -2) \). 2. Intersection of \( x + y = 0 \) and \( x + 3y - 4 = 0 \): \[ \begin{cases} x + y = 0 \\ x + 3y = 4 \end{cases} \] Subtract the first equation from the second: \[ 2y = 4 \Rightarrow y = 2 \] Substitute \( y = 2 \) into \( x + y = 0 \): \[ x + 2 = 0 \Rightarrow x = -2 \] Intersection point: \( (-2, 2) \). 3. Intersection of \( 3x + y - 4 = 0 \) and \( x + 3y - 4 = 0 \): \[ \begin{cases} 3x + y = 4 \\ x + 3y = 4 \end{cases} \] Multiply the first equation by 3: \[ 9x + 3y = 12 \] Subtract the second equation: \[ 8x = 8 \Rightarrow x = 1 \] Substitute \( x = 1 \) into \( 3x + y = 4 \): \[ 3(1) + y = 4 \Rightarrow y = 1 \] Intersection point: \( (1, 1) \). Step 2: Calculate the Lengths of the Sides Using the distance formula \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), calculate the lengths of the sides of the triangle. 1. Length between \( (2, -2) \) and \( (-2, 2) \): \[ \sqrt{(-2 - 2)^2 + (2 - (-2))^2} = \sqrt{(-4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \] 2. Length between \( (2, -2) \) and \( (1, 1) \): \[ \sqrt{(1 - 2)^2 + (1 - (-2))^2} = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \] 3. Length between \( (-2, 2) \) and \( (1, 1) \): \[ \sqrt{(1 - (-2))^2 + (1 - 2)^2} = \sqrt{(3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \] Step 3: Analyze the Triangle - Two sides are equal (\( \sqrt{10} \)), so the triangle is isosceles. - To check if it is right-angled, use the Pythagorean theorem: \[ (4\sqrt{2})^2 = (\sqrt{10})^2 + (\sqrt{10})^2 \] \[ 32 = 10 + 10 \] \[ 32 = 20 \] This is not true, so the triangle is not right-angled. Step 4: Conclusion The triangle is isosceles. Correct Answer: \[ \boxed{C} \]