# Proofs for parallel and perpendicular lines

Does anyone know a good proof to show that all lines parallel to a specific line AB have the same slope? Also, does anyone know a good proof to show that any line perpendicular to a specific line CD will always result in the product of the slopes being -1?

For example, I've seen similarity used to prove that parallel lines have the same slope but I'm looking for other proofs. http://www.analyzemath.com/line/parallel-slope.html

• It's a legitimate question, and Aretino's answer is probably the best answer, but bear in mind, The entire concept of lines obey a formula y=mx +b and m being the "slope" of the line is based on defining the x-y plane to mirror the classical geometry. If you answer the question how does geometry "transer" to analytic plane, you will find this is darned near a tautology. – fleablood Nov 7 '16 at 21:42

Let $y=mx+q$ and $y=m'x+q'$ be the equations of two lines. The coordinates of their intersection are given by the solutions of the system $$\begin{cases} y=mx+q \\ y=m'x+q' \end{cases}$$ This has no solution if and only if $m=m'$ and $q\ne q'$.
The direction vectors of the lines can be written as $v=(1,m)$ and $v'=(1,m')$. For these vectors to be perpendicular, their dot product must vanish, that is: $1+mm'=0$.
Alternatively, perpendicularity can be defined as the least distance direction from a point to line. Taking a generic point $A=(x_A,y_A)$, the square of its distance from a generic point $B=(x,mx+q)$ on the line is given by $$f(x)=(x-x_A)^2+(mx+q-y_A)^2.$$ By setting $f'(x)=0$ one can get the coordinates of $B$ and check that the slope of line $AB$ is $-1/m$.