Since you have two points for each line, we can find the formula for the unique line (assuming the points are distinct). For the first line, defined by $(x_1,y_1)$ and $(x_2,y_2)$, the slope is "rise-over-run", i.e. $\frac {y_2 - y_1} {x_2 - x_1}$, thus any point $x$ and $y$ on this line must have the same slope, so $\frac {y - y_1} {x - x_1}= \frac {y_2 - y_1} {x_2 - x_1}\implies y=\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1\,.$For the second line, we get the analogous result:
$y=\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,.$
Now we just have to solve these two equations, since we are looking for the point of intersection where the $x$ and $y$ variables for each line are equivalent, we set the $y's$ equal to each other and arrive at the equation:
$\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1=\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,,$ which we can simplify to $\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]x= \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,.$
Now if $\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]\ne 0$ we can divide to find a specific formula for $x$: $x=\frac 1 {\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)}\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\right]\,.$ Now, finding $y$ will be easy - we just "back substitute" $x$ into either equation for $y$, say the first one:
$y=\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)\left (\frac 1 {\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)}\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\right] \right) - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1\,.$
This will always work, unless we have that $\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]=0$. In that case, you either have coincident lines or parallel lines - to determine which, just check the y-intercept of each line (we already know the slopes are equal because of the above condition). If the y-intercepts of each line are equal to each other, then the lines are coincident (so there are infinitely many "points of intersection"). If not, they are parallel (and there are no points of intersection).