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

$y = 3x$; $k = \pm 2$

The correct answer is Option (1) → $y = 3x$; $k = \pm 2$ ##

Let $P(x, y)$ be any point on $AB$. Then, area of triangle $ABP$ is zero. So

$\frac{1}{2} \begin{vmatrix} 0 & 0 & 1 \\ 1 & 3 & 1 \\ x & y & 1 \end{vmatrix} = 0$

This gives $\frac{1}{2}(y - 3x) = 0$ or $y = 3x$, which is the equation of required line $AB$.

Also, since the area of the triangle $ABD$ is $3 \text{ sq. units}$, we have

$\frac{1}{2} \begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3$

This gives, $\frac{-3k}{2} = \pm 3$, i.e., $k = \pm 2$.