0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – fleablood Nov 7 '16 at 21:42
0
$\begingroup$

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 $-1/m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.