Find the equation of the line joining $(3, 1)$ and $(9, 3)$ using determinants.

If $A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix}$ is a singular matrix,then the value of $k$ is:

If $D = \left| \begin{array}{ccc} \frac{1}{z} & \frac{1}{z} & -\frac{(x+y)}{z^2} \\ -\frac{(y+z)}{x^2} & \frac{1}{x} & \frac{1}{x} \\ -\frac{y(y+z)}{x^2z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{array} \right|$,then the incorrect statement is:

If ${\left| {\begin{array}{cc} 4 & 1 \\ 2 & 1 \end{array}} \right|^2} = \left| {\begin{array}{cc} 3 & 2 \\ 1 & x \end{array}} \right| - \left| {\begin{array}{cc} x & 3 \\ -2 & 1 \end{array}} \right|$,then $x =$

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(-2, 0), (0, 4), (0, k)$.

$f(x)$ is an $n^{\text{th}}$ degree polynomial satisfying $f(x) = \frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x}) - f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix}$. If $f(2) = 33$,then the value of $f(3)$ is