Three lines are defined by the three equations: 

$$x+y=0$$

$$2x-y=0$$

$$x+2y=2$$

The three lines form a triangle with vertices at:

First, let's write all our equations in a “y=” form:

1. $x+y = 0 \to y = -x$
2. $2x - y = 0 \to y = 2x$
3. $x + 2y = 2 \to y = 1 - \frac{x}{2}$

Each line will intersect with the other two at one point to form a triangle. By finding points where the lines intersect, or are mathematically equal, will give us the three vertices.

Let's take the first line, $y = -x$, and see where it intersects with the two other lines by setting them equal to each other:

$$\begin{gathered} -x = 2x \to x = 0 \\ -x = 1 - \frac{x}{2} \to \frac{-x}{2}=1\to x=2 \end{gathered}$$

Now let's find the corresponding y-coordinates at these two points:

$$\begin{gathered} y = -x \to y = -\left(0\right) \to y = 0 \\ y = -x \to y = -\left(2\right) \to y = -2 \end{gathered}$$

Therefore, we have two of three of our vertices are at (0,0) and (2,-2). To find the last of the three vertices, we must find  where the lines $y = 2x$ and $y = 1-\frac{x}{2}$ intersect:

$$\begin{gathered} 2x = 1-\frac{x}{2}\to \frac{5x}{2}=1\to x = \frac{2}{5} \\ y = 2x \to y = 2\left(\frac{2}{5}\right)\to y = \frac{4}{5} \end{gathered}$$

Our last point is at (2/5,4/5).

Our three vertices are located at: $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{\left(}\raisebox{1ex}{$\mathbf{2}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{5}\mathbf{,}$}\right.\raisebox{1ex}{$\mathbf{4}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{5}$}\right.\mathbf{\right)}$