B2. Determine the coefficient $a$ so that the lines given by the equations $2 x+a y+3=0$ and $3 x-2 y-2=0$ form an angle of $45^{\circ}$.
B2. We write the equations of the lines in explicit form: $y=-\frac{2 x}{a}-\frac{3}{a}$ and $y=\frac{3 x}{2}-1$. The slope coefficients are $k_{1}=\frac{-2}{a}$ and $k_{2}=\frac{3}{2}$. We use the relationship $\tan \alpha=\left|\frac{k_{2}-k_{1}}{1+k_{1} \cdot k_{2}}\right|$ and obtain the equation $\left|\frac{3 a+4}{2 a-6}\right|=1$. From $\frac{3 a+4}{2 a-6}=1$ and $\frac{3 a+4}{2 a-6}=-1$ we get $a_{1}=-10$ or $a_{2}=\frac{2}{5}$.

Writing the coefficients $k_{1}=\frac{-2}{a}, k_{2}=\frac{3}{2}$

$1+1$ point

Writing or using $\left|\frac{k_{2}-k_{1}}{1+k_{1} \cdot k_{2}}\right|=\tan \alpha$ 1 point

Removing the absolute value and writing the equations: $\frac{3 a+4}{2 a-6}=1, \frac{3 a+4}{2 a-6}=-1$. 1 point

Result: $a_{1}=-10, a_{2}=\frac{2}{5}$

$1+1$ point