The intersection of three lines $$x - y = 0$$, $$x + 2y = 3$$ and $$2x + y = 6$$ is a/an

We find the vertices of the triangle formed by the three lines by computing their pairwise intersections.

Intersection of $$x - y = 0$$ and $$x + 2y = 3$$: substituting $$x = y$$ gives $$3y = 3$$, so $$y = 1, x = 1$$. Vertex $$A = (1, 1)$$.

Intersection of $$x - y = 0$$ and $$2x + y = 6$$: substituting $$x = y$$ gives $$3x = 6$$, so $$x = 2, y = 2$$. Vertex $$B = (2, 2)$$.

Intersection of $$x + 2y = 3$$ and $$2x + y = 6$$: from the first equation $$x = 3 - 2y$$, substituting gives $$6 - 4y + y = 6$$, so $$y = 0, x = 3$$. Vertex $$C = (3, 0)$$.

Now we compute the side lengths. $$AB = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}$$. $$AC = \sqrt{(3-1)^2 + (0-1)^2} = \sqrt{5}$$. $$BC = \sqrt{(3-2)^2 + (0-2)^2} = \sqrt{5}$$.

Since $$AC = BC = \sqrt{5} \neq AB = \sqrt{2}$$, the triangle has exactly two equal sides. Therefore it is an isosceles triangle.