How to calculate the length of a line which is perpendicular to another knowing three coordinates? I am sure this is something completely dumb but my mathematics is, well awful, so be kind...
I know three coordinates (2D standard x and y things where the top left is 0,0 and the x increases from left to right and y increases from top to bottom) and I can plot a line between two of them and then make a third point where the intersection with the line should be 90 degrees and what I want to calculate is the length of the line that made this angle (I am sure if I could explain myself properly I would have found an answer to this already).
Here is a picture of what I mean (with some example values for the three coordinates A, B and C that are known to me):

So, how do I calculate the length marked L in the above?
I thought, well, the line L is normal to the vector A to B so I could say...
The vector from A to B is (4, 7) and therefore the normal vectors would be (-7, 4) and (7, -4) but then I am stuck - where do I go now?  Am I even on the right track?
 A: Here is the easiest way I can think of:


*

*Find the normal vector for $AB$.

*Take the dot product of the normal vector and $A$.

*Take the dot product of the normal vector and $C$.

*Subtract the two dot products.

*Divide by the length of the normal vector.
The result is the length of $L$.
Per your example:


*

*$N = B-A = (7, -4)$.

*$a = N \cdot A = (1 \cdot 7) + (1 \cdot -4) = 7 - 4 = 3$.

*$c = N \cdot C = (7 \cdot 7) + (-4 \cdot 5) = 49 - 20 = 29$.

*$c - a = 29 - 3 = 26$.

*$|N| = \sqrt{7^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65} \approx 8.062$.

*$26 \div 8.062 \approx 3.225$.
So $L$ is about 3.225 units long.
A: What you want a point-line distance in 2D.
Let's calculate the formula of $AB$:
$$
y-y_A = \frac{y_B-y_A}{x_B-x_A} (x-x_A)
$$
Plug and simplify:
$$
7x -4y -3 = 0
$$
Now let's use the point-line distance formula:
\begin{align*}
d &= \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}} \\
&= \frac{|7\times7+(-4)\times5+(-3)|}{\sqrt{7^2+(-4)^2}} \\
&= \frac{26}{\sqrt{65}} \approx 3.224
\end{align*}
