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I realize that calculating the midpoint between $2$ points is simple. However, is there a formula to calculate any point between 2 points, like say the point that is $\frac 34$ of the way between 2 points or $\frac 23$ of the way? Right now I am only concerned with the $y$ value and not the $x$ since $x$ is constant.

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    The answer depends a lot on the rules that you want the formula to follow. One can do it "analytically," as suggested by @RSerrao, but if something more primitive, like a circle-and-straightedge construction, is needed, then the answer gets more complicated. – 2017-01-13

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Yes there is! The method is quite simple: think of a straight line between the two points, a line you will walk on. The point that is $T $% along the way is the point on that line, that you reach after walking $T$% of that line's length

You want to go from $A $ to $B $. Start by finding the "arrow" $v $ pointing from $A $ to $B $, like so:

$$v = B - A $$

Then if you want the path that is $\frac{p}{q} $ along the way, do

$$P = A + \frac {p}{q}v = A + \frac{p}{q}(B - A) $$

Actually the function

$$f(t) = A + t (B-A),\ t \in [0, 1] $$

Gives you the point that is $100t$% along the way, given that $f $ happens to be a function that parametrises the line segment from $A $ to $B $

A function $f $ built like that works for $A, B \in \Bbb R^k $, so it need not be in 2 dimensions.

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You can develop the method yourself most easily if you try this with points on a "number line" (i.e. one dimension) before you try extending it two two dimensions or multiple dimensions.

It helps to have an understanding of averages and weighted averages first.

Draw a number line. Choose two points (numbers) on the number line. Find the point directly between them. Now work out the formula that will give you the number (coordinate) of that point, from the other two points. (As you say, it's not hard.)

Now consider a weighted average. If you want your answer to be twice as close to point $A$ as point $B$, then give point $A$ twice as much weight.

If you give point $A$ infinitely more "weight" than point $B$, the result will just be point $A$.

You can extend these ideas easily to two dimensions by considering that there are two separate questions involved: one for the $x$ coordinates, considered as points $A$ and $B$ on a number line, and a different question for the $y$ coordinates.

If you do all of that (which is very simple stuff to work through), you will have begun to develop, on your own, a subject which is called Linear Algebra.