Assertion: The equation of the line joining A(1, 3) and B(0, 0) is given by y=3x.

 Reason: The area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) in the form of determinant is 

$\mathrm{\Delta }=\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

Assume $\mathrm{P}(\mathrm{x},\mathrm{y})$ be any point on AB. Then, area of $△\mathrm{ABP}$ is zero. [since, the three points are collinear]

$\therefore \frac{1}{2}\left|\begin{array}{lll}1& 3& 1\\ 0& 0& 1\\ \mathrm{x}& \mathrm{y}& 1\end{array}\right|=0$

This gives  $\frac{1}{2}(3\mathrm{x}-\mathrm{y})=0$

or y = 3x

Which is the equation of the required line AB.

Reason: The area of triangle with vertices $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$and $\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$ is given by

$\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{lll}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

Therefore, A is true but R is false.