Calculate Michaelis-Menten constant of enzyme catalyzed reaction
What you have given is:
$$\begin{array}{c|c} 
\hline
\dfrac{v}{[S]}, \ \pu{s^{-1}} &amp; v, \ \pu{mol dm^-3 s^-1} \\ \hline
0.257  &amp; 5.15 \times 10^{-2}\\
0.895 &amp; 4.48 \times 10^{-2}\\ 
2.00  &amp; 3.35 \times 10^{-2}\\ 
3.59  &amp; 1.8 \times 10^{-2}\\ 
4.82  &amp; 0.48 \times 10^{-2}\\ 
\hline
\end{array}$$
Michaelis-Menten equation for enzyme kinetics is:
$$v = \frac{V_\mathrm{max} \cdot [S]}{K_M + [S]} \tag1$$
When rearrange this equation (reciprocal) to get Lineweaver-Burk relationship:
$$\frac{1}{v} = \frac{1}{V_\mathrm{max}} + \frac{K_M}{V_\mathrm{max}[S]} \tag2$$
When multiply the equation $(2)$ by $v \times V_\mathrm{max}$, you get:
$$V_\mathrm{max} = v + \frac{vK_M}{[S]} \tag3$$
When rearrange the equation $(3)$, you get $y = mx + c$ type equation (straight-line equation):
$$v  = - K_M \cdot \frac{v}{[S]} + V_\mathrm{max} \tag4$$
The data you have is $\frac{v}{[S]}$ and $v$, which fix the relationship given in the equation $(4)$. Thus, if you can plot them against each other ($\frac{v}{[S]}$ versus $v$) you'd get a straight line with negative lope. The numerical value of the slope is $K_M$ so you can calculate K$_M$ graphically. Also, the $y$-intercept of the graph is equal to $V_\mathrm{max}$:

From the equation of the graph: $\text{|The slope|} = 0.0102$ and $y\text{-Intercept} = 0.054$
Thus, $K_M = \pu{0.0102 mol L-1}$ and $V_\mathrm{max} = \pu{0.054 mol L-1 s-1}$
Once I found OP's mistake of plotting, I was curious to see how this set of data would show the Lineweaver-Burk relationship. Thus, I make the table for $\frac{1}{[S]}$ and $\frac{1}{v}$:
$$\begin{array}{c|c} 
\hline
\dfrac{1}{[S]}, \ \pu{L mol-1} &amp; \dfrac{1}{v}, \ \pu{L s mol-1} \\ \hline
4.99  &amp; 19.42 \times 10^{-2}\\
19.98 &amp; 22.32 \times 10^{-2}\\ 
59.70  &amp; 29.85 \times 10^{-2}\\ 
199.44  &amp; 55.56 \times 10^{-2}\\ 
1004.17  &amp; 208.33 \times 10^{-2}\\ 
\hline
\end{array}$$
As predicted by  Lineweaver-Burk, the plot of $\frac{1}{[S]}$ versus $\frac{1}{v}$ is a straight-line with a positive slope:

From the equation $(2)$, which is the equation of the graph: $\text{The slope} = 0.1891 = \frac{K_M}{V_\mathrm{max}}$ and $y\text{-Intercept} = 18.379 = \frac{1}{V_\mathrm{max}}$
Thus, $V_\mathrm{max} = \frac{1}{18.379} = \pu{0.0544 mol L-1 s-1}$ and $K_M = 0.1891 \times V_\mathrm{max} = 0.1891 \times 0.0544 = \pu{0.0103 mol L-1}$
Both graph give the same values for $K_M$ and $V_\mathrm{max}$.