Find the equation of the line joining $\mathrm{A}(1{,}3)$ and $\mathrm{B}(0{,}0)$ using determinants and find $k$ if $\mathrm{D}(k, 0)$ is a point such that area of triangle $\mathrm{ABD}$ is 3 sq units.

Step 1: Equation of line using determinants

To find the equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ using determinants, we consider a general point $(x, y)$ on the line. The area of the triangle formed by these three collinear points must be zero. The formula for the area of a triangle using determinants is given by the expression shown.

Step 2: Substitute points and solve for equation

Substitute the coordinates of points A $(1{,}3)$ and B $(0{,}0)$ into the determinant formula. Expanding the determinant along the first row gives us $x(3 \cdot 1 - 0 \cdot 1) - y(1 \cdot 1 - 0 \cdot 1) + 1(1 \cdot 0 - 3 \cdot 0) = 0$. Simplifying this expression yields the equation of the line.

Step 3: Area of triangle ABD

Now, we need to find the value of $k$ such that the area of triangle ABD is 3 square units. We use the same determinant formula for the area of a triangle, substituting the coordinates of points A $(1{,}3)$, B $(0{,}0)$, and D $(k,0)$.

Step 4: Substitute points and solve for k

Substitute the coordinates of A $(1{,}3)$, B $(0{,}0)$, and D $(k,0)$ into the area formula. Expanding the determinant along the first row gives $1(0 \cdot 1 - 0 \cdot 1) - 3(0 \cdot 1 - k \cdot 1) + 1(0 \cdot 0 - 0 \cdot k)$. This simplifies to $\frac{1}{2} |-3(-k)| = 3$, which further simplifies to $\frac{1}{2} |3k| = 3$. Solving for $k$ gives two possible values, $k=2$ or $k=-2$.

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