2
$\begingroup$

I am going through one of the most basic line drawing algorithms and stuck with the following mathematical explanations. The implicit function of line equation is:

f(x,y) = ax + by + c

The Book(Computer Graphics , Principles and practice) mentions that

f(x,y) = 0 , when any point m is on line
f(x,y) < 0, when any point m is above the line, and 
f(x,y) > 0, when any point m is below the line.

It would be great to have some explanation of the claim above. I tried to figure out the first one with the following example:

3x + 2y = 1
=> 3x + 2y -1 = 0, where a = 3, b = 2 and c = -1

All I figure out that I need to plug in a (x,y) coordinate so that 3x + 2y - 1 = 0. I am not sure how to choose this coordinate value .

And I am clueless about the next two cases. An example demonstrating all the three cases would be fantastic !

Thanks

$\endgroup$
  • 1
    $\begingroup$ I am unsure what you are asking here. What you posted as quote from the book is just the definition of a line, your title sounds unrelated. Are you asking how to find point on a line of which you have an implicit form like the one you posted? $\endgroup$ – cifz May 30 '16 at 21:59
  • $\begingroup$ The midpoint algorithm uses the implicit equation of the line and I am looking forward to some proofs of the three cases mentioned in my initial post. $\endgroup$ – sajis997 May 30 '16 at 23:51
  • $\begingroup$ This question should be on Math StackExchange. $\endgroup$ – A---B May 31 '16 at 4:51
1
$\begingroup$

enter image description here

(I know i marked the coordinates of the second point wrong)

Let $ax + by + c$ be any line such that $b > 0$.

Now pick point $(\alpha,\beta)$ on the line.

Thus $\alpha *a + \beta*b + c = 0$

Now pick a point $(\alpha, \gamma)$ such that $\gamma$ > $\beta$

Thus $\alpha * a + \gamma * b + c$ > 0

Thus any point satisfying $ax+by +c$ > 0 line above the line.

your results can be deduced from the converse of the proof.

$\endgroup$

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.