Q3: $k$ when the triangle $(-3, 0)$⁠, $(3, 0)$⁠, $(0, k)$ has area 9

The area of the triangle with vertices $(x_1, y_1)$⁠, $(x_2, y_2)$⁠, $(x_3, y_3)$ is the absolute value of

Each row holds one vertex, with a 1 in the last column.

Column 2 has two zeros, so only the entry $k$ contributes.

$k$ is in row 3, column 2, so its sign is $(-1)^{3 + 2} = -1$⁠. Its minor comes from deleting row 3 and column 2:

Now multiply the three parts:

Watch out: $\Delta$ can come out negative, and the area is its absolute value. So both $3k = 9$ and $3k = -9$ are possible.

Check: the base from $(-3, 0)$ to $(3, 0)$ is 6, and the height is $\lvert k\rvert = 3$⁠. Area $= \frac{1}{2} \cdot 6 \cdot 3 = 9$⁠.