Plot a graph of l/l against R

The x-axis will represent $R$ (Resistance in $\Omega$), and the y-axis will represent $\frac{1}{I}$ (Inverse of Current in $\text{A}^{-1}$).
The data points are:

• $(R, \frac{1}{I})$
• $(2 \, \Omega, 5.00 \, \text{A}^{-1})$
• $(4 \, \Omega, 5.60 \, \text{A}^{-1})$
• $(6 \, \Omega, 10.00 \, \text{A}^{-1})$
• $(7 \, \Omega, 12.50 \, \text{A}^{-1})$
• $(9 \, \Omega, 16.70 \, \text{A}^{-1})$

Draw two perpendicular axes on graph paper.

• Label the horizontal axis "$R \, (\Omega)$". Choose a scale from $0 \, \Omega$ to $10 \, \Omega$, with suitable increments (e.g., $1 \, \Omega$ per major grid line).
• Label the vertical axis "$\frac{1}{I} \, (\text{A}^{-1})$". Choose a scale from $0 \, \text{A}^{-1}$ to $20 \, \text{A}^{-1}$, with suitable increments (e.g., $2 \, \text{A}^{-1}$ per major grid line).

Carefully mark each of the five $(R, \frac{1}{I})$ data points on your graph according to their coordinates.

Draw a single straight line that best represents the trend of the plotted points. This line should pass as close as possible to all points, with an approximately equal number of points above and below it. Extend the line to intersect the y-axis.

The resulting graph will be a straight line with a positive slope.