Finding the slope from the given points as:
$$
\begin{aligned}
& \text { Slope }=\frac{y_2-y_1}{x_2-x_1} \\
& \text { Slope }=\frac{3-1}{1-\left(-\frac{1}{2}\right)} \\
& \text { Slope }=\frac{2}{1+\frac{1}{2}} \\
& \text { Slope }=\frac{2}{\frac{3}{2}}
\end{aligned}
$$


Slope $=\frac{4}{3}$
Now the equation will be:
$$
\begin{aligned}
& y=m x+c \\
& y=\frac{4}{3} x+c
\end{aligned}
$$
Now substituting the point $(1,3)$ into the above equation as:
$$
\begin{gathered}
3=\frac{4}{3} \times 1+c \\
3-\frac{4}{3}=c \\
\frac{5}{3}=c
\end{gathered}
$$



So, the equation will be:
$$
\begin{aligned}
& y=\frac{4}{3} x+c \\
& y=\frac{4}{3} x+\frac{5}{3}
\end{aligned}
$$
Now to find the $x$-intercept substitute $y=0$ as:
$$
y=\frac{4}{3} x+\frac{5}{3}
$$


$$
0=\frac{4}{3} x+\frac{5}{3}
$$


$$
\begin{gathered}
\text { undêfined } \frac{20}{3} \underline{2} \text { (20 Sep Shift 2) } \\
x=-\frac{5}{3} \times \frac{3}{4} \\
x=-\frac{5}{4}
\end{gathered}
$$
Therefore, the answer is option [2].