Hello, squat4speed!
What's the orthocenter of a triangle with coordinates: $\displaystyle R (1 , 4) ,\:P (1 , 5) ,\: Q (7 , 2).$
The orthocenter is the intersection of the three altitudes of the triangle.
Find the equation of two of the altitudes and locate their intersection.
If you make a sketch, the problem is much simpler. Code:
P 
(1,5)* 
* * 
* *
*  *
*  *
*  *
  *  +    *Q
*  * (7,2)
*  *
*+*
* *
* * 
(1,4)* 
R 

Side $\displaystyle PR$ is vertical.
Hence, the altitude from $\displaystyle Q$ is the horizontal line, $\displaystyle y = 2.$ .[1]
Side QR has slope: .$\displaystyle m \:=\:\frac{2(\text{}4)}{7(\text{}1)} \:=\:\frac{3}{4}$
Hence, the altitude from $\displaystyle P(\text{}1,5)$ has slope $\displaystyle \text{}\tfrac{4}{3}$
Its equation is: .$\displaystyle y  5 \:=\:\text{}\tfrac{4}{3}(x+1) \quad\Rightarrow\quad y \:=\:\text{}\tfrac{4}{3}x + \tfrac{11}{3}$ .[2]
The intersection of [1] and [2]: .$\displaystyle 2 \:=\:\text{}\tfrac{4}{3}x + \tfrac{11}{3} \quad\Rightarrow\quad \tfrac{4}{3}x \:=\:\tfrac{5}{3}$
Therefore: .$\displaystyle x \:=\:\tfrac{5}{4},\;y \;=\:2$
The orthocenter is: .$\displaystyle \left(\tfrac{5}{4},\:2\right)$