Find intersection of 2 linear equations I have two tables with data, both describing linear equations

Table 1:
x = 60, y = 0
x = 61, y = 1
x = 62, y = 2
x = 63, y = 3
x = 64, y = 4
x = 65, y = 5
x = 66, y = 6
x = 67, y = 7
x = 68, y = 8
x = 69, y = 9
x = 70, y = 10

Table 2:
x = 64, y = 0
x = 65, y = 2
x = 66, y = 4
x = 67, y = 6
x = 68, y = 8
x = 69, y = 10

Part 1:
In $y=mx+b$ for each table, I want to know which is the $m$, the $x$, and the $b$
Part 2: 
I want to find the point where these two lines intersect. The intersection is clear to see by just looking at the data, but I would like the answer to be solved for.
Thanks!
 A: 
Part 1:

For Table 1, we use the fact that the slope is the change in $y$ divided by the change in $x$. Or,$$m=\frac {y_2-y_1}{x_2-x_1}\tag1$$
Randomly choosing two points, say $(70,10)$ and $(69,9)$ and plugging them in, we get the slope as$$m=\frac {10-9}{70-69}=1$$
Thus, we now have $y=x+b$ for $b$ is the $y$-intercept. To find $b$, we plug in any point. Doing so, we get $b=-60$. Therefore, the values in Table $1$ can be represented by $\boxed{y=x-60}$.

For Table 2, we do the same thing that we did in table $1$ to get $\boxed{y=2x-128}$. (See if you can work through it!)


Part 2:

To find the intersection point, you need to find a point $(x,y)$ that satisfies both equations. Therefore, we have the system of equation$$\begin{align*} & y=x-60\\ & y=2x-128\end{align*}\tag2$$
Substituting $y$ with $x-60$, we get the solutions as\begin{align*} & x-60=2x-128\\ & \implies 68=x\end{align*}
Plugging $x=68$ into the first equation of $(2)$ gets us $y$ as $8$. Thus, their intersection point is $\boxed{(68,8)}$
