Okay, I did it the way you suggested. I got the equation of the line to be y=1.5x-5.5
Three of the points are on the line and the fourth isn't far off.
Hi again. As an estimate, I think your equation is fine. I don't know what course you're taking or the intent of the exercise; perhaps, it's a lead into a more advanced discussion of the meaning of line-of-fit and specific method(s). Or, maybe it's just for hands-on fun, for a cursory discussion.
Here's some additional information, if you're interested. Using regression software, I obtained
y = 1.2x - 3.6
That software's goal was to reduce average error. Here's a diagram showing the scatterplot and the graph of the line, onto which I've drawn the error at each point (in red). In this context, 'error' means the vertical distance between the given point and the corresponding point on the line. (In other words, it's the difference between the true y-value and the line-of-fit's y-value.) When the actual point lies above the line of fit, the error is positive, and the error is negative when the actual point lies below the line.
If we average these four signed errors, we get 0. In other words, that's the best line of fit, if we want to minimize the average error between the data points and the line.
The line y = 1.5x - 5.5 yields an average error of 0.5 because there is no error for the first three points (they each lie directly on the line) and the vertical difference between the fourth point and the line is 2.
There are various mathematical methods for finding a line of best fit, given a set of data points. Different methods may generate different equations. In fact, one of those other methods could very well yield your equation, so, without knowing the real purpose of the exercise, I think you did a good job.
Cheers ?
