2. The lines have equations $(1-a) x-2 a y-2=0$ and $-2 x+a y-1=0$. Determine $a$ so that the lines intersect on the bisector of the odd quadrants.
2. If $a=0$, the lines are parallel to each other (their equations are then $x-2=0$ and $2x+1=0$), so we assume that $a \neq 0$. We express $y=\frac{(1-a)x-2}{2a}$ from the first and $y=\frac{2x+1}{a}$ from the second equation. We equate the right sides obtained: $\frac{(1-a)x-2}{2a}=\frac{2x+1}{a}$ and rearrange the equation to $(-a-3)x=4$, from which we express $x=-\frac{4}{a+3}$, if $a \neq -3$ (we can verify that the given equations represent two parallel lines if we consider $a=-3$). We also express the ordinate of the intersection point: $y=\frac{a-5}{a(a+3)}$. Each point on the bisector of the odd quadrants has an abscissa equal to the ordinate, so it must hold that $-\frac{4}{a+3}=\frac{a-5}{a(a+3)}$, from which we finally get $a=1$.

Equation of the bisector $y=x$ 1 point Conversion to a system of two equations with two unknowns 1 point

Correct solving of the system of two equations with two unknowns ... 3 points Solution $a=1$ 1 point