The \(x\)-intercept of a line passing through the points \(\left(\frac{-1}{2}, 1\right)\) and \(B(1,3)\) is

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].